VCE Physics Unit 3: Electronics & Photonics Base notes

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
C
Then 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
Connecting
Wires
Resistor (consumes
energy)
Battery
Current
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

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
Voltmeter
Circuit 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 Revision
Question 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).

Question Type:
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 = I2
R = 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
t
339
-339
0
V2
t
1.15 x 105
0 0
Mean V2
5.8 x 104
t
0
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” (ρ)
(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 sectional
area of Wire 2
∴ Wire 1 has TWICE the resistance
of Wire 2
Combining these mathematically:
R = ρL/A
where:
R = Resistance (Ohms) Ω
ρ = Resistivity (Ohm.m) Ω.m
L = Length (m)

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
R1
R2 R3
V
V1
V2 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 I2
I3
The greater the number of resistors in a series network the greater the
value 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

Question Type:
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

Question Type:
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 2
has a lower (but still constant)
Resistance when
compared to Device 1.
Slope = R
Device 1
Slope = R
Device 2
Georg Ohm

Question Type:
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

Question Type:
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
R1
V1
R2
V2
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 Circuit
VOUT 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 EVERYWHERE
In 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

Question Type:
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.

Question Type:
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 VRMS
VOUT
A B
C
D E
50 Ω
50 Ω
150 Ω
Figure 4
(R1)
(R2)

Question Type:
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.
VOUT
VIN
(R1 + R2)R1
= 6
5000
=
30
(5000 + R2)
R2 = 20,000 Ω = 20kΩ

Question Type:
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
R1
100 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 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.
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 Ω and 700 Ω in series.
I
R2
V2
V
R1
V17 V
5 V 500 Ω
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 Ω,
Then RT = (1/500 + 1/5000)-1
= 454.5 Ω 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 Ω
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 Ω
∴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
R2
V2
V
R1
V1
500 Ω
700 Ω
12
7 V
5 V RL
50 Ω5 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 Ω 5000 Ω
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 single
microprocessor
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 Ph ot on ic s
Photonics 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 Phototransistors
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 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 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
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 Devices
An op amp (operational amplifier)
is a high gain, linear, DC amplifier
The inputs marked as (+) and (-)
do not refer to power supply
connections but instead refer to
inverting and non inverting
capabilities of the amplifier.

Question Type:
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

Question Type:
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

Question Type:
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
digita
l
signal
analogu
e
signal
laser
photodiod
e
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

Question 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.

Question Type:
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 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.
“Modulated” signals consist of
2 components
(a) A carrier signal
(b) An information or data
signal
The carrier is usually of one frequency
and the wave (usually a sine wave) is
y(t) = A sin (ft + φ)
Where
A = Amplitude
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
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 optic
cable, using light.
Question Type:
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 transmit
an audio signal.
Question 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 transmit
an audio signal.
Question 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 Transducers
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.
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

Question Type:
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 converted
A. 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

Question Type:
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) with
changes in light intensity i.e. an
illumination of 105
lux produces a
resistance of 102
ohms.
Question Type:
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 Thermistors
A 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 (o
C)
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

Question Type:
A thermistor is a device the
resistance of which varies with
temperature. The resistance-
temperature characteristic
for 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 200
C.
1000 Ω
Thermistors

Question Type:
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 50
C, 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
Question Type:
Q22. From the list of components below (A–D) select
the one that would be most suitable for use in the
circuit shown
in 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. transistor
D. 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 7
Topics covered:
Amplifiers
Amplifier Gain
Clipping

7.0 Transistor Amplifiers
Shown opposite is a
single stage common
emitter amplifier.
+V
0 V
R1
R2VIN
VOUT
Single stage because
it has only 1 transistor
Common 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
C2
RE
RL
Single stage NPN Transistor
Common Emitter Amplifier
The gain of the
amplifier can be
calculated from:
Gain = VOUT/VIN

Question Type:
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

7.2 Clipping
+V
0 V
R1
R2VIN
VOUT
C1
C2
RE
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
V
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

Question Type:
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)

VCE Physics Unit 3: Electronics & Photonics Base notes

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VCE Physics Unit 3: Electronics & Photonics Base notes