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VCE - PHYSICS

UNIT 3

TOPIC 2

ELECTRONICS & PHOTONICS

TOPIC NOTES

Drouin Secondary College VCE PHYSICS TOPIC: Electronics

Unit Outline

• apply the concepts of current, resistance, potential difference (voltage drop), power to the operation of electronic circuits comprising diodes, resistors, thermistors, and photonic transducers including light dependent resistors (LDR), photodiodes and light emitting diodes (LED); V = IR, P = VI

• calculate the effective resistance of circuits comprising parallel and series resistance and unloaded voltage dividers;

• describe energy transfers and transformations in opto-electronic devices• describe the transfer of information in analogue form (not including the technical aspects of

modulation and demodulation) using– Light intensity modulation i.e. changing the intensity of the carrier wave to replicate the

amplitude variation of the information signal so that the signal may propagate more efficiently– Demodulation i.e. the separation of the information signal from the carrier wave

• design, investigate and analyse circuits for particular purposes using technical specifications related to potential difference (voltage drop), current, resistance, power, temperature, and illumination for electronic components such as diodes, resistors, thermistors, light dependent resistors (LDR), photodiodes and light emitting diodes (LED);

• analyse voltage characteristics of amplifiers including linear voltage gain (ΔVOUT/ΔVIN) and clipping;• identify safe and responsible practices when conducting investigations involving electrical, electronic

and photonic equipment

____________________________________

CHAPTER 1

1.0 Electric Charge

The fundamental unit of electrical charge is that carried by the electron (& the

proton).

This is the smallest discrete charge known to exist independently and is called

the __________________ _______________

Electric Charge (symbol Q) is measured in units called COULOMBS (C).

The electron carries - 1.6 x 10-19 C.

The proton carries +1.6 x 10-19 C.

If 1 electron carries 1.6 x 10-19 C

Then the number of electrons in 1 Coulomb of Charge = 1C/(1.6 x 10-19) =

___________________

1.1 Flowing Charges

When electric charges (in particular electrons) are made to move or “______”,

an Electric Current (symbol I) is said to exist.

VU3E&P Notes


The SIZE of this current depends upon the ___________ OF COULOMBS of

charge passing a given point in a given TIME.

Mathematically: I = Q/t where: I = Current in Amperes (A) Q = Charge in Coulombs (C) t = Time in Seconds (s)

If 1 Amp of current is flowing past this point,

__________________________________________________________

1.2 Electric Current

Electric CURRENTS usually flow along wires made from some kind of

________________ MATERIAL, usually, but not always, a METAL.

Currents can also flow through a Liquid (electrolysis), through a Vacuum (old

style radio “valves”), or through a Semiconductor (Modern Diodes or

Transistors).

A Current can only flow around a ________________ CIRCUIT.

A break ANYWHERE in the circuit means the current stops flowing

EVERYWHERE, IMMEDIATLY.

The current does not get weaker as it flows around the circuit, BUT REMAINS

________________.

It is the ENERGY possessed by the electrons (obtained from the battery or

power supply) which gets used up as the electrons move around the circuit.

In circuits, currents are measured with __________________, which are

connected in series with the power supply.

VU3E&P Notes

Typical Electric Circuit

ConnectingWires

Resistor (consumes energy)

BatteryCurrent

AMeasures

Current

Flow


1.3 Conventional vs Electron Current

Well before the discovery of the ____________, electric currents were known

to exist. It was thought that these currents were made up of a stream of positive

particles and their direction of movement constituted the direction of current

flow around a circuit. This meant that in a Direct Current (D.C.) circuit, the

current would flow out of the _____________ terminal of the power supply and

into the NEGATIVE terminal.

Currents of this kind are called _______________ Currents, and ALL

CURRENTS SHOWN ON ALL CIRCUIT DIAGRAMS EVERYWHERE are

shown as Conventional Current, as opposed to the “real” or ELECTRON

CURRENT.

1.4 Voltage

To make a current flow around a circuit, a _____________ _____________ is

required.

This driving force is the DIFFERENCE in VOLTAGE (Voltage Drop or Potential

Difference) between the start and the end of the circuit.

The larger the current needed, the larger the voltage required to drive that

current.

VOLTAGE is DEFINED as the ENERGY SUPPLIED TO THE CHARGE

CARRIERS FOR THEM TO DO THEIR JOB, ie.TRAVEL ONCE AROUND

THE CIRCUIT.

VU3E&P Notes


Mathematically; V = W/q

where: V = Voltage (Volts) W = Electrical Energy (Joules) q = Charge (Coulombs)

So, in passing through a Voltage of 1 Volt, 1 Coulomb of Charge picks up 1

_______ of Electrical Energy.

OR

A ______ Volt battery will supply each Coulomb of Charge passing through it

with _____ J of Energy.

1.5 EMF Voltage is measured with a ___________________.

Voltmeters are placed in __________________ with

the device whose voltage is being measured.

Voltmeters have a very high internal resistance,

so they have little or no effect the operation of

the circuit to which they are attached.

The term EMF (ELECTROMOTIVE FORCE) describes a particular type of

voltage.

It is the VOLTAGE of a battery or power supply when _______ CURRENT is

being drawn.

This is called the “Open Circuit Voltage” of the battery or supply.

QuestionsQ1: Which one of the following statements (A to D) concerning the voltage across the resistor in Figure 1 is true?

A. The potential at point A is higher than at point B.B. The potential at point A is the same as at point B.C. The potential at point A is lower than at point B.D. The potential at point A varies in sign with time compared to that at point B.

VU3E&P Notes

V

VoltmeterCircuit Symbol


_____________________________________________________________

1.6 Electrical Energy Electrical Energy (W) is defined as the product of the Voltage (V) across, times

the Charge (Q), passing through a circuit element (eg. a light globe).

Mathematically

W = VQ ………1,

where: W = Electrical energy (Joule)V = Voltage (Volts) Q = Charge (Coulomb)

Current and Charge are related through:

Q = It.

substituting for Q, in equation 1 we get:

W = VIt

The conversion of Electrical Energy when a current passes through a circuit

element (a computer) is shown below.

______________________________________________________________Questions

Q2: Determine the electrical energy dissipated in the 100 Ω resistor of Figure 1 in 1 second. In your answer provide the unit.

VU3E&P Notes

Q Coulombs of Electricity enter

computer

Q Coulombs of Electricity leave

computer

In time t, W units of energy are transformed to heat and light

Voltage = V volts

Charges (Q) enter with high energy

Charges (Q) leave with low energy

I I


1.7 Electrical Power

Electrical Power is DEFINED as the Time Rate of Energy Transfer:

Mathematically: P = W/t

where: P = Power (Watts, W) W = Electrical Energy (Joule) t = Time (sec)

From W = VIt we get:

P = VI

From Ohm’s Law (V = IR) [see next chapter] we get:

P = VI = I2R = V2/R

where: I = Current (Amps) R = Resistance (Ohms) V = Voltage (Volts)

Electrical Power is sold to consumers in units of ____________ ________.

(kW.h)

A 1000 W (1kW) fan heater operating for 1 Hour consumes 1kWh of electrical

power.

Since P = W/t or W = P x t, we can say:

1 Joule = 1 Watt.sec

so

1000 J = 1kW.sec

so

3,600,000 J = 1 kW.hour

or

______ MJ = ____ kW.h

1.8 A.C. Electricity

There are two basic types of current electricity:

(a) D.C. (Direct Current) electricity where the current flows in one direction only.

(b) A.C. (Alternating Current) where the current changes ________________ in

a regular and periodic fashion.

The Electricity Grid supplies domestic and industrial users with A.C. electricity.

A.C. is favoured because: VU3E&P Notes


it is cheap and easy to generate

it can be “transformed”; its voltage can be raised or lowered at will by passage

through a transformer.

The only large scale use of high voltage D.C. electricity is in public transport,

i.e. trams and trains.

A.C. ELECTRICITY - PROPERTIES

1.9 R.M.S. Voltage and Current

With an A.C. supply, the average values for both voltage and current = 0,

so Vav and Iav cannot be used by the Power Companies to calculate the amount

of electric power consumed by its customers.

Yet, AC circuits do consume power, so a method of calculating it had to be

found.

To get around this problem R.M.S. or _______ ________ _________ values

for AC voltage and current were developed.

RMS values are DEFINED as:

The AC Voltage/Current which delivers the same voltage/current to an

electrical device as a numerically equal D.C. supply would deliver

An AC source operating at 240 V RMS delivers the same power to a device

as a DC source of 240 V.

VU3E&P Notes

Voltage

Time

VPtoP

T

VP = “Peak Voltage” for Domestic Supply VP = 339 V

VPtoP = “Peak to Peak Voltage”for Domestic Supply VPtoP = 678 V

T = “Period”for Domestic Supply T = 0.02 sec

t

339

-339

0

V2

t

339

0

Mean V2

t

339

0

Mean V2

t339

0

V


1.10 Peak versus RMS Values In AC supplies, the Peak and RMS values are related through simple

formulae:

For Voltage:

VRMS = VP/Ö2

For Current:

IRMS = IP/Ö2

In Australia Domestic Electricity is supplied at 240 V, 50 Hz

The Voltage quoted is the RMS value for the AC supply.

Thus the Peak value for voltage is

VP = VRMS x Ö2

= 240 x 1.414

= 339 V

CHAPTER 2

2.0 Resistance

Electrical Resistance is a property of ________ materials, whether they be

classed as conductors, insulators or something in between. (ie

Semiconductors)

The size of the resistance depends upon a number of factors:

(a) The nature of the material. This is measured by “resistivity” (r)

(b) The ________________, L, of the material.

(c) The _________ ___________________area, A, of the material.

Combining these mathematically:

R = rL/A

VU3E&P Notes

COMPARING RESISTANCE

L


where:

R = Resistance (Ohms) W

r = Resistivity (Ohm.m) W.m

L = Length (m)

A = Cross Sectional Area (m2)

Wires 1 and 2 are made from the same material

Wire 1 has ½ the cross sectional area of Wire 2

\ Wire 1 has ___________ the resistance of Wire 2

2.1 Resistors in Series

Conductors which exhibit a resistance to current flow are generally called

__________________.

When connected “end to end” or in “____________”, the total resistance of the

combination = the sum of the individual resistances of the resistors in the

“network”.

Mathematically: RT = R1 + R2 + R3 + … …

IN A SERIES CIRCUIT:

(a) Since only ________ pathway around the circuit exists, the current through

each resistor is the same. Thus: I = I1 = I2 = I3

b) The sum of the voltage drops across the resistors = the voltage of the power

supply,

Thus: V = V1 + V2 + V3

The greater the number of resistors in a series network the greater the value of the

equivalent resistance (RT)

=

2.2 Resistors in Parallel

VU3E&P Notes

A 2

A 1

R1 R2 R3 RT


Resistors connected “side by side” are said to be connected in

“____________________”.

The total resistance of a parallel network is found from adding the reciprocals of

the individual resistances.

Mathematically: 1/RT = 1/R1 + 1/R2 + 1/R3

IN A PARALLEL CIRCUIT:

(a)The current through each arm varies.

Thus: I = I1 + I2 + I3

(b)The voltage drop across each arm is the same.

Thus: V = V1 = V2 = V3

The greater the number of resistors in a parallel network the lower the value of

the equivalent resistance (RT).

______________________________________________________________

Questions

You wire up the circuit shown in Figure 1 but only have 10 kΩ resistors to work

with.

Q3: Explain how you would construct the R1 = 5 kΩ resistor using only 10 kΩ resistors. Include a sketch to show the connections between the appropriate number of 10 kΩ resistors.

VU3E&P Notes

VIN

VOUT

R2

R1

R3

R2

R1


Q4: Which one of the following statements (A to D) concerning the RMS currents in the circuit of Figure 2 is true?A. The current in resistor A is identical to the current in resistor C.B. The current in resistor D is twice the current in resistor C.C. The current in resistor B is twice the current in resistor E.D. The current in resistor A is identical to the current in resistor D.

2.3 Ohm’s Law

OHM’S LAW relates the __________ across, the ________ through and the

_____________ of a conductor.

Mathematically: V = IR

where: V = Voltage (Volts) I = Current (Amps) R = Resistance (Ohms)

Any conductor which follows Ohm’s Law is called an _____________

CONDUCTOR.

A graph of V versus I produces a

straight line with Slope = ________________

(Remember a straight line

VU3E&P Notes

20 VRMS

VOUT

A B

C

D E

V

I


graph has formula y = mx + c)

The graph is a straight line, \ the Resistance of Device 1 is

CONSTANT (over the range of values studied).

The slope indicates Device 2 has a lesser (but still constant)

Resistance when compared to Device 1.

______________________________________________________________

QuestionsFigure 1 shows a resistor, a linear circuit component, with resistance R = 100 Ω.A DC current, I = 40 mA, passes through this resistor in the direction shown by the arrows.

Q5: What is the voltage drop across this resistor? Express your answer in volts.

2.4 Non Ohmic Devices

Electrical devices which follow Ohm’s Law (V = IR) are called Ohmic Devices.

Electrical devices which do not follow Ohm’s Law are called Non Ohmic

Devices.

Non Ohmics show non linear behaviour when a plot of V vs I is produced, as

can be seen in the graphs for components X and Y opposite.

VU3E&P Notes


Most of the individual components covered in this electronics course are Non

Ohmic Devices.

________________________________________________________________________

QuestionsA resistor is a linear device. An example of a non-linear device is a light-emitting diode (LED).Q6. On the axes provided, sketch a typical current-voltage characteristic curve for each of the devices mentioned.In both cases label the axes and indicate appropriate units.

2.5 Voltage Dividers - 1 Suppose you have a 12 V battery, but you need only 4 V to power your circuit.

How do you get around this problem ?

You use a __________________ __________________ Circuit.

They are made by using combinations of fixed value resistors or using variable

resistors called rheostats.

Voltage dividers are one of the most important circuits types used in

electronics. Almost all sensor subsystems (eg Thermistors, LDR’s), use voltage

divider circuits, there is just no other way to convert the sensor inputs into

useful “electrical” information.

VU3E&P Notes

Current (A)

Voltage (V)

0510

15

1 2 43


For the circuit shown:

V = V1 + V2

Since this is a series circuit ,

the current (I ) is the same everywhere:

I = V1/R1 and I = V2/R2

So V1/V2 = R1/R2

2.6 Voltage Dividers - 2

Using rheostats, the a voltage divider can be set up as shown.

If the main voltage supply (V) is connected across the ends of the rheostat,

then the voltage required by RL is tapped between A and the position of the

slider.

2.7 Voltage Divider Formula

The Voltage divider circuit is a ________________ circuit.

Thus, the SAME CURRENT flows _________________

In other words, the SAME CURRENT flows

through R1 AND R2

VU3E&P Notes

R1V1

R2

V2

2

V

I

V

A

Rheostat

RL

Slider

The further from A the slider movesthe larger the voltage drop across the load resistor , RL

R1

R2

VIN

VOUT

I


THE VOLTAGE DIVIDER FORMULA:

For the VIN circuit:

VIN = I (R1 + R2)\I = VIN / (R1 + R2)………1

For the VOUT circuit:VOUT = IR2

\ I = VOUT /R2………….2Combining 1 and 2 we get:VOUT = VINR2 (R1 + R2)______________________________________________________________

Questions

In Figure 2, five identical 100 Ω resistors are used to construct a voltage divider. The voltage source across this voltage divider is an AC supply with an RMS voltage of 20 V. The resistors are labelled by the letters A to E as shown.Q7: What is the RMS output voltage, VOUT?

VU3E&P Notes

20 VRMS

VOUT

A B

C

D E


An essential component in some of the practical circuits covered in this exam paper is the voltage divider. A DC voltage divider circuit is shown in Figure 1.For the circuit of Figure 1, VIN = 30 V, R1 = 5 kΩ and the output voltage VOUT = 6 V.

Q8: What is the value of the resistance R2? Show your working.

In Figure 1 the 30 V DC input to the voltage divider is replaced by a 100 mV (peak-to-peak) sinusoidal AC input voltage. The resistance values are now R1 = 5 kΩ and R2 = 15 kΩ.

VU3E&P Notes

VIN

VOUT

R2

R1


Q9: What is the current through resistor R2? Show your working, and express your answer as a peak-to-peak current in μA.

2.8 Impedance Matching 1

IMPEDANCE is the TOTAL resistance to current flow due to ALL the

components in a circuit.

In Voltage Divider circuits we only have resistors, so Total Impedance = Total

Resistance.

In the circuit shown a supply of 12 V is connected across 2 resistors of 500 W

and 700 W in series.

The current (I) in the circuit is:

I = V/RT

= 12/1200

= 0.01 A.

The Voltage Drop across R1

= I x R1

= 0.01 x 700

= 7.0 V

The Voltage Drop across R2

VU3E&P Notes

VIN

VOUT

R2

R1

100 mV 5k

15k

I

R2

2

V2

V

R1

1V1


= I x R2

= 0.01 x 500

= 5.0 V

Suppose a load (RL), requires 5.0 V to operate. Conveniently, 5 V appears

across R2.

Lets look at 2 cases where the impedance of RL varies.

CASE (a):

Suppose RL has a total impedance of 50 W

RL and R2 are in parallel,

so Total Resistance RT for the parallel network = (1/R2 + 1/RL)-1

= (1/500 + 1/50)-1

= 45.5 W

\I = V/RT

= 5.0/45.5

= 0.11 A.

This is an 110% increase in the current through R1.

This will cause a dangerous heating effect in R1 and also decrease V across

RL - both undesirable events !

CASE (b): Now RL = 5000 W,

Then RT = (1/500 + 1/5000)-1

= 454.5 W and

I = V/RT

= 0.011 A.

This is only a 10 % increase in current.

In other words it is important to “match” the impedance of the load RL to that of

resistor R2 such that: RL ³ 10R2

Chapter 3

3.0 Semiconductors VU3E&P Notes


• Most electronic devices, eg. diodes, thermistors, LED’s and transistors

are “solid state semiconductor” devices.

• “Solid State” because they are made up of solid materials and have no

____________ parts.

• “Semiconductor” because these materials fall roughly in the middle of the

range between Pure Conductor and Pure Insulator.

• Semiconductors are usually made from ___________ or Germanium with

______________________ deliberately added to their crystal structures.

• The impurities either add extra electrons to the lattice (n type) or create a

deficit of electrons (called “__________”) in the lattice (p type).

• Holes are regarded as positive (+) charge carriers

3.1 p-n junctions Joining together p type and n type material produces a so called

“__________________”

When brought together, electrons from the n type migrate to fill holes in the p

type material.

As a result, a “depletion layer”, (an insulating region containing very few current

carriers), is set up between the two materials.

VU3E&P Notes

p n

depletion layer


The “majority” current carriers are holes in p type material and electrons in n

type material. However, each also has some “______________” carriers

(electrons in p, holes in n) due to impurities in the semiconductor and their

dopeants.

Note: undoped semiconductor material, pure silicon or germanium, is called

“intrinsic semiconductor material”.

3.2 Forward and Reverse Bias If an external supply is now connected as shown it draws the charge carriers

toward the junction and makes the depletion layer smaller.

The current carriers now have enough energy to cross the junction which now

becomes “conducting” or “____________ biased”

If the external supply is now reversed,

it draws the charge carriers away from the junction and makes the depletion

layer bigger meaning current is even less likely to flow and the junction is now

“____________ biased”

3.3 The Diode Diodes are electronic devices made by sandwiching together n type and p type

semiconductor materials.

VU3E&P Notes

p n

depletion layer

p n

depletion layer


This produces a device that has a low resistance to current flow in one

direction, but a high resistance in the other direction.

The “Characteristic Curve” (the I vs V graph) for a typical silicon diode is

shown.

This diode will not fully conduct until a forward bias voltage of 0.7 V exists across it.Notice that when the diode is reverse biased it does still conduct - but the

current is in the pA or μA range. This current is due to _______________

carriers crossing what is for them a forward biased junction.

3.4 The Transistor

There are two general groups of transistors:

• BJT (Bipolar Junction Transistors)

• FET (Field Effect Transistors)

There are two basic types of BJT’s:

• NPN Transistors

VU3E&P Notes

Current (mA)

Voltage (V)


• PNP Transistors

Lets look at the Construction of a BJT npn type transistor

On the circuit symbol the arrow points in the direction of conventional current

flow

Note: npn transistors have the arrow:

Not Pointing iN

3.5 Transistor Uses

The term '_______________' comes from the phrase 'transfer-resistor' because

of the way its input current controls its output resistance.

Transistors are used to perform three basic functions. They can operate as

either

(a)a switch; or (b) an amplifier; or (c) an oscillator

There are over 50 million transistors on a single microprocessor chip. (The

Intel® Pentium 4 has 55 million transistors)

VU3E&P Notes

Collector

Base

Base

Collector

Emitter

Circuit symbol

Emitter

Collector

Base

N

P

N


This is first ever solid state amplifier (transistor) and was created in 1947 at Bell Labs in the US

Chapter 4 Opto Electronic Devices 4.0 Photonics

Photonics is the technology of using light to transmit “________________” from

one place to another.

The light source used is almost always the ___________ and the means of

transmission is the __________ ____________.

Light has the ability to transmit “information” at a much faster rate than electrons in

copper wires.

Photonics main use is in telecommunications. With optical fibres costing only a

fraction of previously used copper wires and having the ability to carry far more

information, telecommunications has been revolutionised by the use of photonics.

Photonic devices fall into 2 general categories: Photovoltaic they generate their

own ____________ and do not require an external power supply, example solar

cells,

Photoconductive require an external supply and operate by modifying the

_________________, example would be a Light Dependent Resistor (LDR) or

Photodiode

4.1 Photodiodes

Photodiodes are detectors containing a p-n semiconductor junction.

They are unique in that they are the only device that can take an external

stimulus and convert it directly to _________________.

Photodiodes are commonly used in circuits in which there is a load resistance

in series with the detector.

The output is read as a change in the voltage drop across the resistor.

VU3E&P Notes


The magnitude of the photocurrent generated by a photodiode is dependent

upon the ____________________ of the incident light. Silicon photodiodes

respond to radiation from the ultraviolet through the visible and into the near

infrared part of the E-M spectrum.

The photovoltaic detector may operate without external bias voltage.

A good example is the solar cell used on spacecraft and satellites to convert

the sun’s light into useful electrical power.

4.2 Phototransistors

Phototransistors are used extensively to detect _________ pulses and convert

them into digital electrical signals. In an optical fibre network these signals can

be used directly by computers or converted into analogue voice signals in a

telephone.

Like diodes, all transistors are light-sensitive.

Phototransistors are designed specifically to take advantage of this fact.

The most-common variant is an NPN bipolar transistor with an exposed base

region. Here, light striking the ___________ replaces what would ordinarily be

voltage applied to the base -- so, a phototransistor amplifies variations in the

light striking it. Phototransistors may or may not have a base lead (if they do,

the base lead allows you to bias the phototransistor's light response.

VU3E&P Notes

RL

VO

UT

+V

0 V


Note that photodiodes also can provide a similar function, although with much

lower gain (i.e., photodiodes allow much less current to flow than do

phototransistors).

4.3 Phototransistor Applications Phototransistors can be used as light activated switches.

Further applications1. Optoisolator- the optical equivalent of an electrical transformer. There is no

physical connection between input and output.

2. Optical Switch – an object is detected when it enters the space between

source and detector.

VU3E&P Notes

RL

+V

0V

VOUT

When light is on - VOUT is High

RL

+V

0V

VOUT

When light is on - VOUT is Low


4.4 Optoisolator Circuit

How does VOUT respond to changes to VIN ?

As the input signal changes, IF changes and the light level of the LED changes.

This causes the base current in the phototransistor to change causing a change

in both IC and hence VOUT

The response of the phototransistor is not instantaneous, there is a lag

between a change in VIN showing up as a change in VOUT

Assume VIN varies such that the LED switches between saturation (full on) and

cut off (full off), producing a square wave variation in IF

IC will respond showing a slight time lag every time IF changes state

VU3E&P Notes

IF

t

IC

t


4.5 Opto-electronic Devices

QuestionsYou are asked to investigate the properties of an optical coupler, sometimes called an opto-isolator. This comprises a light-emitting diode (LED) that converts an electrical signal into light output, and a phototransistor (PT) that converts incident light into an electrical output. Before using an opto-isolator chip you consider typical LED and PT circuits separately.

VU3E&P Notes

An op amp (operational amplifier) is a high gain, linear, DC amplifierThe inputs marked as (+) and (-) do not refer to power supply connections but instead refer to inverting and non inverting capabilities of the amplifier.


A simple LED circuit is shown in Figure 4 along with the LED current-voltage characteristics. The light output increases as the forward current, IF , through the LED increases.

Q10: Using the information in Figure 4, what is the value of the resistance, RD, in series with the LED that will ensure the forward current through the LED is IF

= 10 mA?

Q11: Will the light output of the LED increase or decrease if the value of RD is a little lower than the value you have calculated in the last question? Justify your answer.

VU3E&P Notes


You now consider the phototransistor (PT) circuit of Figure 5 with RC = 2.2 kΩ. The light is incident upon the base region of the PT and produces a collector current, IC.Q12: As the light intensity incident on the PT increases, which one of the following statements concerning the PT-circuit of Figure 5 is correct?A. The collector current remains constant, but VOUT increases.B. The collector current remains constant, but VOUT decreases.C. The collector current increases, but VOUT decreases.D. The collector current decreases and VOUT decreases.

4.6 CD Readers

Compact discs store information in ________ form. This information is

extracted by a laser and photodiode combination. The data is passed through a

series of electronic processes to emerge from the speaker as sound

Questions

VU3E&P Notes

CD pits

DAC

amplifier


The information on an audio CD is represented by a series of pits (small depressions) in the surface that are scanned by laser light. When there is no pit the reflected light gives a maximum light intensity, I1, detected by a photodiode circuit. When the laser light strikes a pit, the light intensity is reduced to I0. A plot of a typical light intensity incident on the photodiode is shown in Figure 4.The variation in current as a function of light intensity for the photodiode is shown in Figure 5a, together with the circuit used to determine this, which is shown in Figure 5b.

Q13: With no light incident upon the photodiode, the current in the photodiode circuit, the “dark current”, is 5 µA.What is the output voltage, VOUT, across the 100 Ω resistor in the circuit of Figure 5b?

Chapter 5 5.0 Analogue Data VU3E&P Notes


The world is divided into two streams: ____________ and ______________

Humans perceive the world as an _______________ place i.e. we receive our

input is a continuous stream, this continuous stream is what defines analogue

data.

On the other hand digital data (a stream of 1’s and 0’s) estimates analogue

data by “__________________” it at various time intervals

Analogue data is usually more ______________________ than digital data.

However _______________ data is easier to store and manipulate and of

course computers can only cope with digital data

Digital systems are not just a modern invention.

Examples of ancient digital systems include:

The Abacus Morse Code Braille Semaphore

5.1 Modulation Modulation is a a way of changing an

analogue signal so data or information

can be transmitted over a

communication network.

The carrier is usually of

one frequency and the

wave (usually a sine wave)

is

y(t) = A sin (ft + φ)

Where

A = Amplitude

VU3E&P Notes


f = Frequency

φ = Phase

Changing (modulating) this wave can only occur by changing one of A, f or φ

Changing A leads to Amplitude Modulation

Changing f leads to Frequency Modulation

Changing φ leads to Phase Modulation

5.2 Demodulation Demodulation is

the inverse process

of modulation. The

modulated wave

signal is

transmitted to a receiver at the receiving station.

Then information components are extracted from the carrier signal (recovering

information).

The process is called demodulation.

5.3 Fibre Optics The idea of using visible ________ as a medium for communication had

occurred to Alexander Graham Bell back in the late 1870s, but he did not have

a way to generate a useful carrier frequency or to transmit the light from point to

point.

All forms of modern communication--radio and television signals, telephone

conversation, computer data--rely on a ___________ signal. By modulating the

carrier, we can encode the information to be transmitted; the _________ the

carrier frequency, the more ___________________ a signal can hold.

In 1960, an idea first introduced by Albert Einstein more than 40 years earlier

bore practical fruit with the invention of the __________.

VU3E&P Notes


This achievement prompted researchers to find a way to make visible light a

communication medium--and a few years later _________ _________ arrived.

Questions

Figure 9 is a sketch of an electro-optical system that allows sound to be transmitted over a distance via a fibre opticcable, using light.Q14. Explain the terms modulation and demodulation as they apply to the transmission of sound by this system.

Figure 8a, below, shows a schematic diagram for an intensity-modulated fibre-optic link that is used to transmitan audio signal.To test the device an audio signal is fed into the microphone. The signal at point W is shown in Figure 8b.Q15. Which of the diagrams (A–D) below best represents the signal observed at point X in Figure 8a?

Q 16. Which of the diagrams (A–D) above could represent the signal that would be observed at point Y in Figure 8a?

VU3E&P Notes


Chapter 6 6.0 Input Transducers

___________________ are devices which convert non electrical signals into

electrical signals. Input Transducers convert mechanical and other forms of

energy eg. Heat, Light or Sound into Electrical Energy.

Examples of such devices are :

Light Emitting Diode (LED)Light is emitted when the diode is forward biased

Light Dependent Resistor (LDR)The resistance changes as light intensity varies

ThermistorThe resistance changes as the temperature changes

PhotodiodeCurrent flows when light of a particular frequency illuminates the diode.

6.1 Light Emitting Diodes LEDs emit light when an electric current passes through them.

LEDs must be connected the correct way round.

The diagram may be labelled a or + for anode and k or - for cathode (yes, it

really is k, not c, for cathode!). The cathode is the short lead and there may be

a slight flat region on the body of round LEDs.

VU3E&P Notes

Circuit Symbol

a k


LEDs must have a _______________ in series to limit the current to a safe

value

Notice this is a ________________ divider circuit

Most LEDs are limited to a maximum current of 30 mA, with typical VL values

varying from 1.7 V for red to 4.5 V for blue

______________________________________________________________

QuestionsThe LED in Figure 4 is an electro-optical converter.

Q17. Which one of the following statements (A to D) regarding energy conversion for the LED is correct?All the electrical energy supplied from the DC power supply is convertedA. only to heat energy in both the resistor, RD, and the LED.B. partly to heat energy in the resistor, RD, the remainder to light-energy output from the LED.C. partly to heat energy in both the resistor, RD, and the LED, with the remainder to light-energy output from the LED.

VU3E&P Notes


D. to heat energy in the LED, with the remainder to light-energy output from the LED.

Q18: Describe the basic purpose of each of the following electronic transducers.i. Light-Emitting Diode (LED)ii. Photodiode

______________________________________________________________

6.2 Light Dependent Resistors (1) The light-sensitive part of the LDR is a wavy track of cadmium sulphide.

Light energy triggers the release of extra charge carriers in this material, so that

its resistance __________ as the level of illumination increases.

A light sensor uses an LDR as part of a voltage divider.

Suppose the LDR has a resistance of 500Ω , (0.5 kΩ), in bright light, and 200

kΩ in the shade (these values are reasonable).

When the LDR is in the light, Vout will be

When the LDR is in the dark, Vout will be:

VU3E&P Notes


In other words, this circuit gives a LOW voltage when the LDR is in the light,

and a HIGH voltage when the LDR is in the shade.

A sensor subsystem which functions like this could be thought of as a

'________ sensor' and could be used to control lighting circuits which are

switched on automatically in the evening.

6.3 Light Dependent Resistors (2) The position of the LDR and the fixed resistor are

now swapped.

How does this change affect the circuit’s operation ?

Remember the LDR has a resistance of 500Ω ,

(0.5 kΩ), in bright light, and 200 kΩ in the shade.

In the light:

In the dark:

This sub system could be thought of as a “__________ sensor” and could be

used to automatically switch off security lighting at sunrise.

QuestionsThe graph opposite shows the variation in resistance of a light dependent resistor (LDR) withchanges in light intensity i.e. an illumination of 105 lux produces a resistance of 102 ohms.

VU3E&P Notes

Vout 10 10 + 0.5 = x 9 = 8.57 V

Vout 10 10 + 200 = x 9 = 0.43 V


Q19. What is the resistance of the LDR when the light level is 103 lux?

6.4 Thermistors A temperature-sensitive resistor is called a ______________.

The resistance of most common types of thermistor _________________ as

the temperature rises.

They are called negative temperature coefficient, or ntc, thermistors. Note the -

t° next to the circuit symbol.

Different types of thermistor are manufactured and each has its own

characteristic pattern of resistance change with temperature.

The diagram shows characteristic curve for one particular thermistor:

Note the ______ scale for resistanceVU3E&P Notes

Resistance (Ω)

Temp (oC) 20 40 60 80

100

1000

10000

100000


6.5 Thermistor Circuits

How could you make a sensor circuit for use as a fire alarm?

You want a circuit which will deliver a ________ voltage when hot conditions

are detected.

At 80o RThermistor = 250 Ω (0.25 kΩ)

How could you make a sensor circuit to detect temperatures less than 4°C to

warn motorists that there may be ice on the road?

You want a circuit which will give a _________ voltage in cold conditions.

At 4o RThermistor = 40 kΩ

VU3E&P Notes

R = 10 k

10 10 + 0.25 = x 9 = 8.78 VVout

R = 10k

40 10 + 40 = x 9 = 7.2 VVout


_____________________________________________________________

Questions

A thermistor is a device the resistance of which varies with temperature. The resistance-temperature characteristicfor a thermistor is shown in Figure 7.

Q20. What is the value of the resistance of the thermistor at 20°C?

The thermistor is incorporated into the control circuit for the refrigeration unit of a cool room. The circuit is shown in Figure 8

The relay switches the refrigeration unit ON when voltage, V, across variable resistor R ≥ 4V and switches OFF when V < 4V.The refrigeration unit must turn on when the temperature of the cool room rises to, or exceeds, 5°C.Q21. At what value should the resistor R be set so that the refrigeration unit turns on at this temperature?You must show your working.

VU3E&P Notes


Figure 9 is a sketch of an electro-optical system that allows sound to be transmitted over a distance via a fibre optic cable, using light.

Q22. From the list of components below (A–D) select the one that would be most suitable for use in the circuit shownin Figure 9 at position P and the one most suitable for use at position Q.A. LDR (light dependent resistor)B. LED (light emitting diode)C. transistorD. diode

CHAPTER 7

7.0 Transistor Amplifiers

Shown below is the single stage common emitter amplifier.

Single stage because it has only 1 transistor

Common emitter because the emitter is common to both input and output.

VU3E&P Notes

+V

0 V

R1

R2VIN

VOUT

RL

RE C2

C1


The device can be regarded as a black box (dotted line) with an input and an

output

The voltage divider consisting of R1and R2 provides the ___________ bias so

the base will be positive with respect to the emitter. Resistors are sized to set

the _______________ (Q) or steady state operating point at the middle of the

load line (shown by the dot on load line see below).

RL is chosen to limit the __________________ current to the maximum allowed

value.

RE is chosen to set VCE at the voltage which will allow the biggest “swing” in the

output signal to occur.

So this amplifier is now correctly biased and can operate to produce an

enlarged (amplified), inverted output.

7.1 Gain

The gain of the amplifier can be calculated from:Gain = VOUT/VIN

QuestionsThe graph of vOUT versus vIN for the transistor amplifier is shown in Figure 4.

VU3E&P Notes


Q23. What is the voltage amplification of the transistor amplifier?You must show your working.

Q24. Explain the shape of the graph in Figure 4. Your explanation should include why the graph shown has a negative slope, and why it has horizontal sections at vIN > +60 mV and vIN < –60 mV.

7.2 Clipping

The load line for an amplifier is a plot of the collector emitter voltage against the

base emitter voltage

VU3E&P Notes


Setting the Q point of the amplifier at an incorrect level can lead to the output

signal being distorted, cut off or “_____________ ”

Trying to drive the amplifier too hard, by having too large an input signal will also lead to clipping of the output signal

QuestionsThe input signal, vIN, she is using for the amplifier mentioned in Q 23 is shown in Figure 5.

Q25. On the graph below, sketch the output signal as measured at point vOUT.

VU3E&P Notes

VCE (V)

VBE (V)

Q

VIN

VOUT

Q set too high – top of signal clipped

Q

VIN

VOUT

Q set too low – bottom of signal clipped


VU3E&P Notes

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