Download - VCE Physics Unit 3 - Topic 2 Electronics & Photonics

VCE Physics

Unit 3 - Topic 2

Electronics &

Photonics

1.0 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

• Topics covered:

• Electric Charge.

• Electric Current.

• Voltage.

• Electromotive Force.

• Electrical Energy.

• Electric Power.

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 ELEMENTARY CHARGE.

• 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 CThen the number of electrons in 1 Coulomb of Charge

= 1 C 1.6 x 10-19 = 6.25 x 1018 electrons

1.1 Flowing Charges• When electric charges (in particular

electrons) are made to move or “flow”, an Electric Current (symbol I) is said to exist.

• The SIZE of this current depends upon the NUMBER OF COULOMBS of charge passing a given point in a given TIME.

Section of Current Carrying Wire

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,

then 6.25 x 1018 electrons pass here every second.

1.2 Electric Current• Electric CURRENTS usually flow along

wires made from some kind of CONDUCTING 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 COMPLETE 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 CONSTANT.

• 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 AMMETERS, which are connected in series with the power supply.

Typical Electric Circuit

ConnectingWires

Resistor (consumes energy)

BatteryCurrent

A

Measures Current Flow

1.3 Conventional Current vs Electron Current

Positive Terminal Negative Terminal

Conventional vs Electron Current

Resistor

Electron Current: Never shown on Circuit Diagrams

Conventional Current:Always shown on Circuit Diagrams

Well before the discovery of the electron, 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 POSITIVE terminal of the power supply and into the NEGATIVE terminal.Currents of this kind are called Conventional 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 DRIVING FORCE 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.

• So, in passing through a Voltage of 1 Volt, 1 Coulomb of Charge picks up 1 Joule of Electrical Energy.

• OR • A 12 Volt battery will supply each

Coulomb of Charge passing through it with 12 J of Energy.

Mathematically;

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

Alessandro Volta

http://www.cinemedia.net/SFCV-RMIT-Annex/rnaughton/VOLTA_BIO.html

1.5 E.M.F. Voltage is measured with a VOLTMETER.

The term EMF (ELECTROMOTIVE FORCE) describes a particular type of voltage.It is the VOLTAGE of a battery or power supply when NO CURRENT is being drawn.

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

V

VoltmeterCircuit Symbol

With S closed, a current begins to flow and V drops and now measures voltage available to drive the current through the external circuit

Resistor

A

V

S

V measures EMF

Voltmeters are placed in PARALLEL 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.

Resistor

A

V

10

Electronics & Photonics RevisionQuestion Type:

Q1: 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.

Potential Difference

1.6 Electrical Energy

The conversion of Electrical Energy when a current passes through a circuit element (a computer) is shown below.

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

Voltage = V volts

Charges (Q) enter with high energy

Charges (Q) leave with low energy

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

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).


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

Electrical Energy

A: Electrical energy W = VQ = VIt = (4.0)(40 x 10-3)(1)

= 0.16 Joule

1.7 Electrical Power• Electrical Power is DEFINED as the

Time Rate of Energy Transfer:

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

• From W = VI t 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 Kilowatt-Hours. (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

3.6 MJ = 1 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 direction in a regular and periodic fashion.

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

• A.C. is favoured because:

(a) it is cheap and easy to generate

(b) 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, ie. trams and trains.

Voltage

Time

VP VPtoP

T

A.C. ELECTRICITY - PROPERTIES

VPtoP = “Peak to Peak Voltage”

for Domestic Supply VPtoP = 678 V

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

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

1.9 R.M.S. Voltage and Current

V

t339

-339

0

V2

t

1.15 x 105

0 0

Mean V2

5.8 x 104

t0

Mean V2

240 t

GRAPHICAL DEVELOPMENT OF THE RMS VOLTAGE FROM AN A.C. VOLTAGE

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.

To get around this problem R.M.S. or Root Mean Square 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.

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

1.10 Peak versus RMS Values• In AC supplies, the Peak

and RMS values are related through simple formulae:

• For Voltage:

VRMS = VP/2

• For Current:

IRMS = IP/2

• 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 2 = 240 x 1.414 = 339 V

Voltage (V)

Time (s)

VP

+339 V

- 339 V

VP to P

240 V

Chapter 2

• Topics covered:

• Resistance.

• Ohm’s Law.

• Resistors in Series and Parallel.

• Voltage Dividers

• Impedance Matching

2.0 Resistance• Electrical Resistance is a property of

ALL 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” ()

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

(c) The cross sectional area, A, of the material.

COMPARING RESISTANCE

L

A 2

A 1

Wires 1 and 2 are made from the same material

Wire 1 has ½ the cross sectionalarea of Wire 2

Wire 1 has TWICE the resistance of Wire 2

Combining these mathematically:

R = L/A where:R = Resistance (Ohms) = Resistivity (Ohm.m) .m L = Length (m) A = Cross Sectional Area (m2)

2.1 Resistors in Series• Conductors which exhibit a

resistance to current flow are generally called RESISTORS.

• When connected “end to end” or in “SERIES”, 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 ONE pathway around the circuit exists, the current through each resistor is the same.

Thus: I = I1 = I2 = I3

Resistors in SERIES

These three resistors can be replaced by a single resistor of value

RT = R1 + R2 + R3

R1R2 R3

V

V1V2 V3

Resistors in a Series Circuit

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

Thus: V = V1 + V2 + V3

I

I1 I2I3

The greater the number of resistors in a series network the greater thevalue of the equivalent resistance (RT)

R1 R2 R3

RT

2.2 Resistors in Parallel• Resistors connected “side by side”

are said to be connected in “PARALLEL”.

• The total resistance of a parallel network is found from adding the reciprocals of the individual resistances.

IN A PARALLEL CIRCUIT:(a) The current through each arm varies.

Thus: I = I1 + I2 + I3

R3

R2

R1

These three Resistors can be replaced by a single Resistor ( RT )

Resistors in Parallel

Resistors in a Parallel Circuit

R3

R2

R1

V

I3

I2

I1

I

V1

V3

V2

(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).

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

RT


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.

VIN

VOUT

R2

R1

Connect two 10k resistors together in parallel

VIN

VOUT

R2

R1

Parallel Resistors


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.

20 VRMS

VOUT

A B

C

D E

Current Flows

2.3 Ohm’s Law• OHM’S LAW relates the Voltage

across, the Current through and the Resistance 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 OHMIC CONDUCTOR.

Ohm’s Law - Graphically

V

I

A graph of V versus I produces a straight line with Slope = R(Remember a straight line 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 2has a lower (but still constant)Resistance when compared to Device 1.

Slope = RDevice 1

Slope = RDevice 2

Georg Ohm


Figure 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. A: V = IR

= (40 x 10-3)(100) = 4.0 V

Ohm’s Law

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.

• Most of the individual components covered in this section of the course are Non Ohmic Devices.

Voltage (V)

Current (A)

Component Y

0

5

10

15

2 4 6 8

Current (A)

Voltage (V) Component X

0

5

10

15

1 2 43


A 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.

I(Amp)

V(volts)

a linear device – a resistor a non linear device – an LED

V(volts)

I(mA)

Ohmic & Non Ohmic Devices

2.5 Voltage Dividers - 1

For the circuit above:

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

R1V1

R2V2

V

I

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 Voltage Divider 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.

2.6 Voltage Dividers - 2

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.

V

A

Rheostat

RL

Slider

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

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

Slider type rheostat

Various rotary rheostats

2.7 Voltage Divider Formula

For the VOUT circuit:

VIN = I (R1 + R2)

VIN CircuitVOUT CircuitR1

R2

VIN

VOUT

I

For the VIN circuit: Applying Ohm’s Law

The Voltage divider circuit is a SERIES circuit.Thus, the SAME CURRENT flows EVERYWHEREIn other words, the SAME CURRENT flows through R1 AND R2

I = VIN

(R1 + R2) …….(1)

VOUT = IR2

I = VOUT

R2

……..(2)

Combining 1 and 2 we get:VOUT = VIN R2 (R1 + R2)

so, VOUT = VIN.R2

(R1 + R2)

This is the Voltage Divider Formula


20 VRMS

VOUT

A B

C

D E

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?

Voltage Divider Network

A: Step 1 Determine the equivalent resistance for A and B and D and E

For A and B: 1/RE = 1/RA + 1/RB = 1/100 + 1/100 = 2/100 So RE = 100/2 = 50 Ω

Replace A and B with one 50 Ω resistor, same with D and E.

You can now redraw the circuit.


The series circuit in Figure 3 can be further simplified as shown in Figure 4

20 VRMS

VOUT

A B

C

D E

50 Ω

50 Ω

100 Ω

Figure 3

VOUT =VIN R2

(R1 + R2)

The original question (value of VOUT) can now be calculated, using the Voltage Divider formula:

= 20(150)

(50 + 150) = 15 V


20 VRMSVOUT

A B

C

D E

50 Ω

50 Ω

150 Ω

Figure 4

(R1)

(R2)


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.

VIN

VOUT

R2

R1

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.


VOUTVIN

(R1 + R2)R1

= 65000

=30

(5000 + R2)R2 = 20,000 Ω = 20kΩ


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.

VIN

VOUT

R2

R1100 mV

5k

15k

A: Series circuit – add resistances so RT = 20kΩ

Use Ohm’s Law to find current V = IR so I = 100 x 10-3

20 x 103= 5μA

Ohm’s Law

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 1IMPEDANCE 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.

The current (I) in the circuit is: I = V/RT = 12/1200 = 0.01 A.

In the circuit shown a supply of 12 V is connected across 2 resistors of 500 and 700 in series.

I

R2V2

V

R1V17 V

5 V 500

700 Ω

12

The Voltage Drop across R1 = I x R1

= 0.01 x 700 = 7.0 V

The Voltage Drop across R2

= I x R2

= 0.01 x 500 = 5.0 V

CASE (b): Now RL = 5000 , Then RT = (1/500 + 1/5000)-1 = 454.5 and I = V/RT = 0.011 A.This is only a 10 % increase in current.

CASE (a): Suppose RL has a total impedance of 50

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 I = V/RT = 5.0/45.5 = 0.11 A. This is an 110% increase in the current in the circuit. This will cause a dangerous heating effect in R1 and also decrease the Voltage across RL - both undesirable events !

Suppose a load (RL), requires 5.0 V to operate.Conveniently, 5 V appears across R2.

2.9 Impedance Matching 2

I

R2V2

V

R1V1

500

700 Ω

12

7 V

5 V RL 50 5 V

In other words it is important to “match” the impedance of the load RL to that of resistor R2 such that: RL 10R2

5 V 500 5000

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

Chapter 3.

• Topics covered:

• Semiconductors

• Diodes

• p-n junctions

• Forward & Reverse Bias

• Capacitors

3.0 Semiconductors• Most electronic devices, eg. diodes,

thermistors, LED’s and transistors are “solid state semi conductor” devices.

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

• “Semiconductor” because these materials fall roughly in the middle of the range between Pure Conductor and Pure Insulator.

• Semiconductors are usually made from Silicon or Germanium with impurities deliberately added to their crystal structures.

• The impurities either add extra electrons to the lattice producing n type semiconductor material.

N - Type Semiconductor

Si Si

Si Si

P Si

Si Si

extra electron

P - Type Semiconductor

Si Si

Si Si

B Si

Si Si

hole

or create a deficit of electrons (called “holes”) in the lattice producing p type semiconductor material.Holes are regarded as positive (+) charge carriers, moving through the lattice by having electrons jump into the hole leaving behind another hole.

3.1 p-n junctions

p n

Joining together p type and n type material produces a so called “p-n junction”

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

np

np

As a result, a “depletion layer”, (an insulating region containing very few current carriers), is set up between the two materials.

depletion layer

The “majority” current carriers are holes in p type material and electrons in n type material.However, each also has some “minority” 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

p n

depletion layer

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 “reverse biased”

p n

depletion layer

it draws the charge carriers toward the junction and makes the depletion layer smaller.

If an external supply is now connected as shown

The current carriers now have enough energy to cross the junction which now becomes “conducting” or “forward biased”

If the external supply is now reversed,

3.3 The Diode• Diodes are electronic devices made by sandwiching together n

type and p type semiconductor materials.• This produces a device that has a low resistance to current flow

in one direction, but a high resistance in the other direction.

Cathode (-)Anode (+)

Conventional Current Flow

Current (mA)

Voltage (V) 0.7 V

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 minority carriers crossing what is for them a forward biased junction.

V (μA)

Circuit Symbol

3.4 The TransistorThere are two general groups of transistors:

•BJT (Bipolar Junction Transistors)•FET (Field Effect Transistors)

There are two basic types of BJT’s:•NPN Transistors•PNP Transistors

Lets look at the Construction of a BJT npn type transistor

Emitter

Collector

Base

Base

Collector

Emitter

Circuit symbol

N

P

N

Note: npn transistors have the arrow: Not Pointing iN

The arrow points in the direction of conventional current flow

An npn type transistor

3.5 Transistor Uses

Transistors are used to perform three basic functions. They can operate as either(a) a switch; or

(b) an amplifier;

There are over 50 million transistors on a singlemicroprocessor chip.(The Intel® Pentium 4 has 55 million transistors)

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

or (c) an oscillator

The term 'transistor' comes from the phrase 'transfer-resistor' because of the way its input current controls its output resistance.

Chapter 4

Topics Covered:

Opto - Electronic Devices

4.0 PhotonicsPhotonics is the technology of using light to transmit “information” from one place to another. The light source used is almost always the laser and the means of transmission is the optical fibre.

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

Photonics main use today 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: Photovoltaics they generate their own voltage and do not require an external power supply, example solar cells.Photoconductive require an external supply and operate by modifying the current, example Light Dependent Resistor (LDR) or Photodiode

4.1 Photodiodes

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.

Photodiodes are detectors containing a p-n semiconductor junction.

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.

The magnitude of the photocurrent generated by a photodiode is dependent upon the wavelength 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.

RL VOUT

+V

0 V

They are unique in that they are the only device that can take an external stimulus and convert it directly to electricity.

4.2 PhototransistorsLike 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 base 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.

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).

Phototransistors are used extensively to detect light 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.

4.3 Phototransistor Applications

RL

+V

0V

VOUT

RL

+V

0V

VOUT

When light is on VOUT is High

When light is on VOUT is Low

Phototransistors can be used as light activated switches.

Further applications

1. 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.

4.4 Optoisolator CircuitHow 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

IF

t

IC

t

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

4.5 Opto-electronic DevicesAn 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.


You 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.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?For 10 mA to flow through LED requires a voltage of 1.5 V (read from graph)Because LED and R are in series VRD = 10 – 1.5 = 8.5 V

VRD = IRD so RD = VRD/I = 8.5/(10 x 10-3) = 850 Ω

Ohm’s Law


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.

Justification: Reducing the value of RD will not affect the voltage drop across it.The Voltage across RD is controlled by the LED which will remain at 1.5 V thus VRD will still equal 8.5 V.So if V remains the same and R goes down I must go up. So a larger current flows through the LED meaning an increased light output

Increased output

Ohm’s Law


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.

Phototransistor

4.6 CD Readers

CD pits

digital

signal

analogue

signal

lase

r photodiode

DAC

digital to analogue converter

amplifier speaker

Compact discs store information in Digital 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

Electronics & Photonics RevisionQuestion Type: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.


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?

Ohm’s Law

A: The photodiode and resistor are in series.The same current flows through each. VOUT = V100Ω = IR = (5 x 10-6)(100)

= 5 x 10-4 V

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.

Chapter 5

Topics Covered:•Analogue Data•Modulation•Demodulation•Optical Fibres

5.0 Analogue Data The world is divided into two streams:Analogue and Digital

Humans perceive the world as an analogue 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 “sampling” it at various time intervals

Analogue data is usually more accurate than digital data. However digital 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 AbacusMorse CodeBrailleSemaphore

5.1 ModulationModulation is a a way of changing an analogue signal so data or information can be transmitted over a communication network. “Modulated” signals consist of

2 components (a) A carrier signal(b) An information or data

signalThe carrier is usually of one frequency and the wave (usually a sine wave) is y(t) = A sin (ft + φ)WhereA = Amplitudef = Frequencyφ = Phase

Changing (modulating) this wave can only occur by changing one of A, f or φChanging A leads to Amplitude ModulationChanging f leads to Frequency ModulationChanging φ leads to Phase Modulation

5.2 DemodulationDemodulation 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

All forms of modern communication--radio and television signals, telephone conversation, computer data--rely on a carrier signal.By modulating the carrier, we can encode the information to be transmitted; the higher the carrier frequency, the more information 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 laser.

The idea of using visible light 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.

This achievement prompted researchers to find a way to make visible light a communication medium--and a few years later fibre optics arrived.

A.G. Bell

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.

In the context of this question, modulation referred to variation of the light intensity. Demodulation meant that the variation in the light intensity created an electrical signal.

Modulation and Demodulation

Figure 8a, below, shows a schematic diagram for an intensity-modulated fibre-optic link that is used to transmitan audio signal.

Electronics & Photonics RevisionQuestion Type: Modulation and Demodulation

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?

At X there is light of varying intensity, so the answer was D. The light intensity cannot go below zero, so B was not an option. Even when there is no signal at W to modulate the output of the laser diode, there is a uniform brightness emitted.

Figure 8a, below, shows a schematic diagram for an intensity-modulated fibre-optic link that is used to transmitan audio signal.

Electronics & Photonics RevisionQuestion Type: Modulation and Demodulation

To test the device an audio signal is fed into the microphone. The signal at point W is shown in Figure 8b.Q 16. Which of the diagrams (A–D) above could represent the signal that would be observed at point Y in Figure 8a?

The varying brightness incident on the photodiode would cause a voltage like C to be produced.

Chapter 6

Topics Covered:

Input Transducers

6.0 Input TransducersTransducers 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.

Light Emitting Diode (LED)

Light is emitted when the diode is forward biased

Light Dependent Resistor (LDR)

The resistance changes as

light intensity varies

Symbol

Examples of a few such devices are shown here.

Photodiodes

Current flows when light of a particular frequency illuminates the diode

Thermistor

The resistance changes as the temperature changes

6.1 Light Emitting Diodes

anode (+)

cathode (-)

flat edge

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.

Circuit Symbol

a k

LEDs must have a resistor in series to limit the current to a safe value

Notice this is a voltage 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


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

Energy Conversion


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

(i) Emits visible light when a current flows through it. Converts electrical energy to light.

(ii) Switches on (allows a current to flow through it) when exposed to light. Converts light to electrical energy.

Transducer Properties

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 falls 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:

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 'dark 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.

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 “light sensor” and could be used to automatically switch off security lighting at sunrise.

How does this change affect the circuit’s operation ?

Vout 10 10 + 0.5

= x 9 = 8.57 V

Vout 10 10 + 200

= x 9 = 0.43 V

The 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.


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

Read from graph R = 104 Ω = 10,000 ohms

LDR

6.4 ThermistorsA temperature-sensitive resistor is called a thermistor. There are several different types:The resistance of most common types of thermistor decreases 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.

Resistance (Ω)

Temp (oC) 20 40 60 80

100

1000

10000

100000

Note the log scale for resistance

The diagram shows characteristic curve for one particular thermistor:

6.5 Thermistor Circuits

R = 10 k

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

At 80o RThermistor = 250 Ω (0.25 kΩ)

10 10 + 0.25

= x 9 = 8.78 VVout

R = 10 k

You want a circuit which will deliver a HIGH voltage when hot conditions are detected.

You need a voltage divider with the ntc thermistor in the position shown:

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 HIGH voltage in cold conditions.

You need a voltage divider with the thermistor in the position shown:

At 4o RThermistor = 40 kΩ

40 10 + 40

= x 9 = 7.2 VVout


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?

From the graph read off the resistance for a temp of 200C. 1000 Ω

Thermistors


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.

At 50C, the thermistor resistance = 4000 ohms. Use the voltage divider relationship to determine the value of R. 4 = 12R / (4000 + R)R = 2000 Ω

Thermistors

Transducers


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 The LED was the best option for P and the LDR was

the best for Q.

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.

Chapter 7Topics covered:AmplifiersAmplifier GainClipping

7.0 Transistor AmplifiersShown opposite is a single stage common emitter amplifier.

+V

0 V

R1

R2VIN

VOUT

Single stage because it has only 1 transistorCommon emitter because the emitter is common to both input and output.

The voltage divider consisting of R1and R2 provides the forward bias so the base will be positive with respect to the emitter. Resistors R1 & R2 are sized to set the quiescent (Q) or steady state operating point at the middle of the load line (shown by the green dot on load line, see next slide).

RL

RE C2

RL is chosen to limit the collector 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.

The device can be regarded as a black box with an input and an output

C1

So this amplifier is now correctly biased and can operate to produce an enlarged (amplified), inverted output.

7.1 Gain+V

0 V

R1

R2VIN

VOUT

C1

C2RE

RL

Single stage NPN Transistor Common Emitter Amplifier

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


The graph of vOUT versus vIN for the transistor amplifier is shown in Figure 4.Q23. What is the voltage amplification of the transistor amplifier?You must show your working.

Voltage amplification (gain) is the magnitude of gradient of the graph. Gain = (3 – 0) /(0- 60 x 10-3) = - 50. + 50 is also acceptable. Q24. Explain the shape of the graph in

Figure 4. Negative slope: An increase in VIN leads to a corresponding decrease in Vout. This is an inverting amplifier

Amplifiers

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.

Horizontal section for VIN > +60 mV: the amplifier is saturated, i.e. maximum current flows through the transistor.Horizontal section for VIN < −60 mV: the amplifier is at cut-off, i.e. minimum (zero) current flows through the transistor.

7.2 Clipping+V

0 V

R1

R2VIN

VOUT

C1

C2RE

RL

Setting the Q point of the amplifier at an incorrect level can lead to the output signal being distorted, cut off or “clipped”

VCE (V)

VBE (V)

Q

VIN

VOUT

Q set too low – bottom of signal clipped

Q

VIN

VOUT

Q set correctly – no clipping

Q

VIN

VOUT

Q set too high – top of signal clipped

Single stage NPN Transistor Common Emitter Amplifier

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

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


The input signal, vIN, she is using is shown in Figure 5.

Q25. On the graph below, sketch the output signal, VOUT.

Amplifiers

A student is studying the performance of the inverting amplifier in question 23 . It has a gain of 50

Clipping occurs because the input signal can only vary between ± 60 mV (see Fig 4 previous question)

The End

Ollie Leitl 2005

c

Download - VCE Physics Unit 3 - Topic 2 Electronics & Photonics

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