Download Energy Transfer (Work Done)

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9 Photonics 2015
• 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);
Current
EMF
Q
I
t
E
V
Q
Energy Transfer (Work Done)
W=VQ
= V I t (as Q = I t)
Power
W
t
VIt
=
t
=VI
P =
V = IR
V
I=
R
Hence, it can be seen that
P  VI
 I2R
V2

R
P, Q and R, are three identical light globes connected to a voltage source, . The light globes
always have the same resistance, regardless of the current flowing through them.
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Example 1
1986 Question 56
(1 mark)
rate of heat dissipated in R
What is the ratio:
?
rate of heat dissipated in Q
Example 2
1986 Question 57
(1 mark)
If globe Q is removed from its socket, what is the ratio:
current flowing through lamp R before lamp Q was removed
?
current flowing through lamp R after lamp Q was removed
Example 3
1986 Question 58
(1 mark)
Which of the following statements (A - D) best describes any change in the potential difference
between points X and Y as a result of removing lamp Q?
A.
The potential difference between points X and Y will increase.
B.
The potential difference between points X and Y will become zero.
C.
The potential difference between points X and Y will not change.
D.
The potential difference between points X and Y will decrease.
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Transducers
A transducer is a device that converts a physical signal to an electrical signal or an electrical signal
to a physical signal.
Input Transducers
An input transducer detects a physical change in the environment near an electrical circuit and
converts it into an electrical signal. The way that we will be using the input transducers is to have
them in a voltage divider situation and by measuring the voltage over the device you can determine
the resistance. You then use this resistance on a graph and determine the temperature or light
intensity. Examples of input transducers would be a microphone, thermistors, LDR, Photodiode
etc.
Output Transducers
An output transducer converts an electrical signal into a form that can be used. Examples of output
transducers would be a speaker, an LED, etc.
Light Dependent Resistor (LDR)
The resistance of a light dependent
resistor (LDR) is determined by the
amount of light that falls on the top of it.
When light shines on the LDR, it has low
resistance and allows current to flow.
When light does not shine on it, the LDR
has a very high resistance, and a much
smaller current will not flow through it.
vout
If you take the ‘voltage out’ over the
Figure 1
LDR, as seen in figure 1, then as the
light intensity increases, the ‘voltage out’
decreases. If you
take the ‘voltage out’ over the other resistor the voltage will increase as the light intensity
increases.
Note logarithmic scales; resistance depends on illumination (lux); lot of light – low resistance; little
light – high resistance; relatively slow to respond (ms).
Light Emitting Diode (LED)
A light emitting diode (LED) is a diode that emits light when a current flows
through it in the forward direction. LED’s are used in electronic displays as
indicator lamps. Energy is dissipated in the diode-either in the form of heat
or by emitting photon, LED’s are designed so that the energy is released
as light.
The arrow in the symbol indicates that it emits light. Figure 6 shows an
LED in the forward biased direction. In this direction the LED will glow. An
LED will not allow current to flow in the reverse direction. LEDs
require a larger voltage to run then the normal diodes. They require
around 2V, depending on the colour of the LED (1.7V for red and 2.2V for
green).
vout
Figure 6
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25mA
5mA
1.6V 1.7V
If the amount of current flowing through the LED is changed, the amount of light being emitted will
also change. This way the LED can be used as an Optical output. The intensity of the light varies
directly with the current flowing through the LED. An LED is not 100% efficient in converting
electrical energy into light. When a LED starts conducting, the resistance drops sharply, as seen
from the graph. With a very small change in voltage, there is a marked increase in current.
Note
No light is emitted until about 1.6V (forward bias voltage)
After this threshold, resistance drops rapidly
LEDs need to be connected in series with a resistor to limit the current through them
LEDs can emit in many colours and infrared
They transform electrical energy into light energy
Much more efficient that incandescent filaments, they switch on much faster and last a lot longer
Figure 1 shows a circuit where a light emitting diode (LED) will emit light, or not, depending on the
current. The circuit consists of the LED, a variable resistor R and a 12 V battery. Figure 2 shows
the current-voltage characteristics of the LED.
Example 4 1999 Question 1
(3 marks)
For values of the variable resistor between 150 Ω and 750 Ω, the LED emits light and the voltage
across it stays the same. Calculate the voltage across the variable resistor when its resistance is
350 Ω.
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Example 5 1999 Question 2
(3 marks)
Calculate the current in the LED when the resistance of the variable resistor is 350 .
A seatbelt warning device uses a circuit similar to that in Figure 1. The switch represents the
seatbelt. When the switch is open it represents an open seatbelt and, when closed, a closed
seatbelt. Figure 3 shows three possible circuits. Notice that in Circuit C the voltage across the LED
is opposite to that in the other two circuits.
Example 6 1999 Question 3
(6 marks)
Circle the letter (A – C) above, showing the circuit which will ensure that the LED emits light only
when the seatbelt is open. Explain why the circuit you have selected is suitable and the other
two are not.
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The resistance of a particular thermistor varies with temperature as shown below. The thermistor is
placed in a circuit that allows the voltage across the thermistor, and the current through it, to be
measured.
When the thermistor is placed in a cup of coffee, the measured voltage across it is 4.0 V and the
current through it is 10 mA.
Example 7 1998 Question 4
(4 marks)
What is the temperature of the coffee? Show your working.
Photodiodes
The current through a
photodiode is directly
proportional to the light
intensity. The brighter the
light the more current
there will be. If there is no
light there will be no
current. Photodiodes have
a very fast reaction rate so
they are used in digital
communications.
vout
Using a photodiode in a circuit.
First work out what the current will be by the amount
of light energy shining on the photodiode. For our
example let’s say it was 2 Wm-2. This means the
current is 10A. Now you multiply this current by R
and find the voltage across R. If the voltage across R
is larger than the voltage supplied then the voltage
across R is the voltage supplied.
The photodiode can be both photovoltaic (generators
of potential difference) and photoconductive
(modifiers of an electric current), depending on the application.
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A reverse-biased photodiode operates in what is called photoconductive mode, since the
conduction varies with the illuminating light intensity. If the reverse-biased voltage is relatively large
(i.e. several volts) the reverse-biased photodiode will have a very fast response time (much faster
than an LDR) and is suitable for detecting light signals that vary down to a time scale of a fraction
of a microsecond.
All diodes and transistors interact with light (and heat) to some degree. Some diodes and
transistors are built to shut out any ambient light and prevent it affecting the operation of the
device, but photonic devices use the incoming light to change the energy levels of the charges.
Depending on the application, these devices operate as a source of current or function by
modifying existing currents.
Photovoltaic mode
Even if there is no external power supply, the junction of two semiconductor materials can act as a
photovoltaic cell. Photovoltaic cells, or photocells, convert light energy into electrical energy
directly and they produce an output current when light falls on them. Solar cells are an example of
this type of device.
Typical current produced in a photodiode ranges from about 1nA (in the dark) to about 100nA (in
bright light). (Note: 1nA = 10-9 A.)
Photodiodes are used in a series combination to produce higher voltage and in parallel to produce
higher current. Photodiodes can also be used to generate an electrical signal or pulse from the
light energy, and tend to be specially made for either energy or signal production.
Photoconductive mode
Photodiodes
When connected to an external power supply the diode acts as a photoconductive device. This is
opposite to the photovoltaic device. The current is often in proportion to the incident light intensity.
Thus one function of a photodiode is to switch and regulate a working current, depending on the
light intensity striking the device. Some devices can be made sensitive to infra-red light by
selecting a casing for the phototransistor that is transparent to infra-red but opaque to visible light.
These devices are used in security systems and in remote control devices for TVs and videos. The
advantage of using infra-red light is that any visible light in the room does not accidentally operate
the device.
Photoconductive devices respond to a change in the intensity of the incident light with a change in
their electrical resistance. This allows more, or less, current to flow in the device. In many
applications the device acts as an on/off switch as the device goes from cut off (or nearly fully off or
very little current) to saturation (or nearly fully on).
Combining ohmic and non-ohmic elements in a circuit.
By using resistors such as a thermistor or a LDR, where the resistance varies with temperature or
light intensity you can obtain values for the temperature or the light intensity by reading the voltage
over the variable resistor, calculating the resistance, then using the characteristic curve (graph) to
find the values.
Remember if the device in connected in series then it must have the same current flowing through
it and you can often work out the voltages from that. If the devices are connected in parallel than
the potential difference across both devices will be the same.
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Circuit Symbols
Light dependent resistor (LDR)
The resistance of a light dependent
resistor (LDR) is determined by the
amount of light that falls on it. Graphs
of resistance and light intensity are
used with this component.
Diode
When a diode is placed in a circuit,
the current only flows through it in one
direction. A silicon diode takes 0.7V.
The diode is used to keep currents
flowing in one direction.
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Thermistor
A thermistor is a temperature sensitive
resistor. Graphs of resistance and
temperature are used with this
component.
Photodiode
This is a reverse biased diode that uses
light to allow the current to flow. These
are used as very quick light detectors.
The current in a photodiode circuit is
determined by the light at the
photodiode.
Light Emitting Diode (LED)
LEDs work the same way as a normal
diode. They only allow current to flow
in one direction. The only difference
is the LED emits light at the same
time. The amount of light that they
emit is dependant on the current
flowing through the LED.
This type of problem has become quite popular over the last few years.
A light dependent resistor (LDR) has the characteristics shown below.
In a measurement of the light intensity in a classroom, a student measures the resistance of the
LDR to be 3000 ohm.
Example 8 2007 Question 7
(2 marks)
What is the intensity of the light falling on the LDR?
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A circuit is set up to switch on a light when the illumination in the room drops below 20 lux. The
light comes on when the voltage across VOUT reaches 4.0 V.
The circuit is shown below
Example 9 2007 Question 8
(2 marks)
What should be the value of the resistor, R, for the circuit to operate correctly?
The class decides that the room is too dark before the light comes on. The teacher wishes to reset
the system so that the light comes on before the classroom becomes so dark.
Example 10 2007 Question 9
(2 marks)
In the space below indicate whether the value of R should be increased or decreased to achieve
this. Explain your answer.
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Example 11 2005 Question 8
(2 marks)
Describe the basic purpose of each of the following electronic transducers.
i. Light-Emitting Diode (LED)
ii. Photodiode
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
below.
The variation in current as a function of light intensity for the photodiode, together with the circuit
used to determine this, are shown below.
Example 12 2005 Question 9
(2 marks)
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?
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A resistor is a linear device. An example of a non-linear device is a light-emitting diode (LED).
Example 13 2005 Question 10
(4 marks)
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.
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A solar cell is a non-linear diode device that is able to provide electrical power when exposed to
light. In the test circuit shown in Figure 2a, the power dissipated in the variable load resistor R
depends on the light intensity, and the value of the R. For this example Figure 2b shows the solar
cell characteristic curve current I versus voltage V for the light intensity used.
Example 14 2001 Question 4
(2 marks)
If R = ∞ ohm, the solar cell provides no current. What is the voltage across the resistor R?
Example 15 2001 Question 5
(2 marks)
If R is decreased to 0 ohm, the voltage across the solar cell approaches zero. What is the
maximum current that the solar cell can deliver to the load resistor R?
Example 16 2001 Question 6
(2 marks)
At what point (A, B, C or D) on the characteristic curve of Figure 2b is the largest electrical power
dissipated in resistor R?
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Part of a-home safety system is designed to draw attention to low light levels. This part uses a light
dependent resistor (LDR), with characteristics as graphed below. Note that the scales are not
linear.
The LDR is connected as part of a voltage divider, as shown below. The resistor has a value of
270k.
Example 17 1992 Question 1
(2 marks)
What is the resistance of the LDR when the light level is 10 lux?
Example 18 1992 Question 2
(2 marks)
When the light level is 10 lux, what is the voltage at point X?
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Modulation
V
Signal
Light
Resistor
The light coming from the LED is modulated (varying) in
time will the signal that you want to send.
Light is an electro-magnetic wave and the intensity of the
light is increased by increasing the amplitude of the
electric field (and magnetic field).
So if we want to use light as our medium to send our
signal on we will be sending our signal on a wave, so the
light is called a carrier wave since it is a wave that
carries the signal.
Electric field
Light Intensity
Below is a diagram of a constant light intensity and what the electric field would look like at that
time.
time
time
Electric field
Light intensity
Now if you want to modulate (vary) the light intensity as shown below the electric field will need to
do the following.
time
time
The signal is now being carried on the carrier wave. There is a need for the carrier wave to have a
frequency much greater than the signal, otherwise it will do a poor job of representing the signal
being sent.
Modulation
This is simply altering the intensity of a beam of light in a way that transfers information
Eg. If the intensity of a beam of light were varied sinusoidally with a frequency of 440Hz, it could
convey the ‘information’ of a frequency of 440Hz
The information of a single 440Hz frequency is not much, however, all sounds, including speech,
can be made up of a combination of such frequencies. Hence it is possible to attach speech
information (or music, or other sounds) to a light beam.
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Demodulation
Light
Demodulation is the process of taking the input light and
converting it back to a signal. For this course we will look at the
Photodiode as the component that will take the light intensity
and convert it back to a signal. If we take the voltage across
the resistor the voltage will correspond to the original signal
that was sent.
V
Resistor
This means that we need to remove the signal from the carrier
wave.
The received signal
The carrier wave looks like the graph shown below
From this we can see that the original signal must have been the one shown below.
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The figure below is a sketch of an electro-optical system that allows sound to be transmitted over a
distance via a fibre-optic cable, using light.
Example 19 2006 Question 8
(4 marks)
Explain the terms “modulation” and “demodulation” as they apply to the transmission of sound by
this system.
Modulation
Demodulation
Example 20 2006 Question 9
(2 marks)
From the list of components below (A–D) select the one that would be most suitable for use in the
circuit shown 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
Energy Transfers and transformations
In photonic applications, energy is often being transformed between electrical potential energy and
light (photon) energy.
LED:
LDRs:
Phototransistors:
Photodiodes;
Incandescent lights:
Electrical potential energy → photon (light energy)
Photon (light) energy → electrical potential energy of electrons in the LDR
Photon energy to electrical potential energy of electrons
Photon energy to electrical potential energy and kinetic energy of electrons
Electrical potential energy of electrons to photon energy and heat.
There is always some energy lost in transference, usually as heat.
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Information
Consider the following block diagram.
Information
source
Sound into
microphone


Converter
(to varying
voltage)
amplifier
Electro-optical
converter
Opto-electrical
converter
LED circuit
Light beam
Photodiode
circuit
Converter
(to varying
voltage)
amplifier
Converter
(voltage to
information)
loudspeaker
The labels underneath the diagram are possible components for such an arrangement.
There are many other possibilities.
A simplified circuit diagram of such an arrangement is shown below:
R1
C
Signal from
information source







LED
R2
to amplifier
The AC signal is derived from the information source, such as a microphone/amplifier
combination
The battery in the left hand circuit adds a DC component to the input to the LED, this
ensures that the light level of the LED is steady. The AC component of the signal causes
this level to fluctuate, thus causing modulation of the intensity.
The resistor R1 protects the LED from too much current ( a standard feature in LED circuits)
The light from the LED is transmitted to the phototransistor in the right hand circuit (using
optical fibres or lenses)
The fluctuating intensity falling on the phototransistor causes the collector current through
the circuit to change in exactly the same way as the signal from the information source.
(linear transducers)
Using V = iR, the voltage is connected to the input of an amplifier (note the blocking
capacitor to ensure that no DC is involved)
The output of the amplifier then reproduces the original information as sound, (using a
speaker) or a waveform (using a CRO)
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You are asked to design a circuit to detect a customer as they enter and leave a shop by walking
through a light beam. The light source is a light-emitting diode (LED), and the light sensor is a lightdependent resistor (LDR). The circuit is shown below.
In order to test this design you first consider the LED circuit by itself; this is shown below along with
the LED current-voltage characteristics
Example 21 2004 Sample Question 7
(2 marks)
What is the current in the circuit shown above?
Show your working and express your answer in mA.
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Having established that this works correctly you now consider the complete circuit, including the
LDR. The characteristics of this device are shown below.
Example 22 2004 Sample Question 8
(3 marks)
When the light beam is not broken the incident light intensity at the LDR is 100 lux. Calculate the
current in the LDR circuit, ILDR. Show your working and express your answer in mA.
Example 23 2004 Sample Question 9
(2 marks)
When the light beam is broken as a customer walks through the door, the voltage measured
across the 900 Ω resistor in series with the LDR is measured as 0.01 V. What is the incident light
intensity at the LDR? Show your working.
Example 24 2004 Sample Question 10
(2 marks)
The light emitting diode (LED) in the figure is an electro-optical converter.
Which one of the following statements (A – D) regarding energy conversion and transformation for
the LED is correct?
A
If the current through the LED increases, the intensity of the light emitted by the LED also
increases even though the voltage across the LED does not alter. All electrical energy is
converted to light energy so it is a 100% efficient conversion process.
B
If the current through the LED increases, the intensity of the light emitted by the LED also
increases even though the voltage across the LED does not alter. Not all electrical energy is
converted to light energy so it is less than a 100% efficient conversion process as some energy
will be dissipated as heat energy.
C
If the current through the LED increases, the intensity of the light emitted by the LED remains
unchanged because the voltage across the LED does not alter. All electrical energy is converted
to light energy so it is a 100% efficient conversion process.
D
If the current through the LED increases, the intensity of the light emitted by the LED remains
unchanged because the voltage across the LED does not alter. Not all electrical energy is
converted to light energy so it is less than a 100% efficient conversion process as some
energy will be dissipated as heat energy.
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An opto-electronic system is used to transmit digital information between computers. In order to
transmit this information on a light beam within an optic fibre, the intensity of the light beam is
modulated by the instantaneous magnitude of the information voltage signal. The figure below shows
the time variation of the voltage signal for some specific information. Note that the voltage has both
positive and negative values.
Example 25 2004 Sample Question 11
(2 marks)
On the graph immediately below the information signal, sketch the light intensity as a function of
time that could convey this information using intensity modulation. At the left-hand side of the
graph the light intensity is indicated when no information (0 V) is being conveyed.
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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 belong along with the LED current-voltage characteristics. The light
output increases as the forward current, IF , through the LED increases.
Example 26 2004 Pilot Question 7
(2 marks)
Using the information above, 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?
Example 27 2004 Pilot Question 8
(2 marks)
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 above? Justify your answer.
Example 28 2004 Pilot Question 9
(2 marks)
The LED in the figure above is an electro-optical converter.
Which one of the following statements (A – 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.
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Solutions
Example 1 1986 Question 56
rate of heat dissipated in R PR
=
rate of heat dissipated in Q PQ
Use P = i2R.
The current in R is twice the current in Q,
(because there is the same current in P and
Q).
(2i)2 R
P
 R =
(i)2 R
PQ
=4
(ANS)
Example 2 1986 Question 57
When the globe Q is removed the effective
resistance of the circuit changes.
Initially the total resistance of the circuit is
given by
Rtotal = R + Rparallel
1
1
1
Rparallel is given by
=
+
Rparallel
R
R
R
2
3R
Rtotal =
2
2V
V
Using I =
we get I =
3R
R
When globe Q is removed the total
resistance becomes R + R = 2R
V
V
Using I =
we get I =
R
2R
Therefore
Rparallel =
current flowing through lamp R before lamp Q was removed
current flowing through lamp R after lamp Q was removed
2V
4
3R
becomes
=
V
3
2R
(ANS)
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Example 3 1986 Question 58
The potential difference between X and Y
(which is the same as across P) changes
because the resistance of the circuit changes.
3R
Initially Rtotal =
, when globe Q was
2
removed Rtotal became 2R.
1
Therefore initially
of the voltage was lost
3
1
across the parallel section, but finally
of
2
the voltage was lost across P (XY)
So the potential drop across XY increases.
A
(ANS)
Example 4 1999 Question 1
(42%)
If the voltage drop across the LED is 1.5V,
then the LED will emit light.
The voltage across the resistor
V = 12V - VLED
= 12 - 1.5 = 10.5V
You need to check that this answer is
possible.
If V = IR, across the resistor then
10.5 = I × 350
I = 10.5  350
I = 0.03 A
I = 30 mA.
(ANS)
This is a suitable answer because the LED
must have the same current as the resistor,
(they are in series), and the LED can easily
have a current of 30 mA, when the voltage
across it is 1.5V.
Example 5 1999 Question 2
(43%)
You should note that good checking of
answers often produces the answer to the
next question.
If V = IR, across the resistor then
10.5 = I × 350
I = 10.5  350
I = 0.03 A
 I = 30 mA (ANS)
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Example 6 1999 Question 3
(51%)
You need to read these questions very
carefully, often the best time to do this is
during your reading time, you will get 15
minutes to read the paper before you are
allowed to write. Use it wisely.
In circuit A, when the switch is open, (i.e. an
open seatbelt), you don't get a complete
circuit, so the LED will not emit light when the
seatbelt is open.
In circuit B, when the switch is open, the
charge will flow through the LED, and when
the switch is closed, (i.e. the seatbelt is
closed (seatbelt buckled up)) then all the
charge will go through the switch, because it
offers a lot less resistance, so the LED will
not emit light.
In circuit C, The LED will not emit light when
the switch is open because the diode will not
allow current to flow in the reverse direction.
B
(ANS)
Example 7 1998 Question 4
You need to use the information to find the
resistance first. (I hope that by this stage of
your preparation you have automatically
highlighted the units on the axes, so that you
realised that you were given a resistance v
temperature graph. This meant that you
needed one of either of these to use the
graph).
Using V = IR 4.0 = 10mA × R
R = 4.0/ 10 × 10-3
R = 400
From the graph, the temperature that
represents a resistance of 400 = 700.
 700 (ANS)
Example 8 2007 Question 7
(62%)
When the LDR has a resistance of 3000Ω or
3kΩ the light that is falling on the LDR is 10
lux, read straight from the graph.
 3000 Ω (ANS)
Example 9 2007 Question 8
(62%)
This is a simple voltage divider question. First
find the resistance of the LDR when it has 20
lux falling on it. This is found to be 1500Ω
read straight from the graph. The LDR has 4V
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across it so the resistor must have 2V across
it.
V
R
Now use 1 = 1 ,
V2 R2
2
 R2 = ×1500 = 750 Ω.
4
 R = 750 Ω (ANS)
Example 10 2007 Question 9
(62%)
If the light is to come on before the intensity is
reduced to 20 lux then the LDR will have a
resistance less than 1500Ω (read from the
graph), so if there is to be 4 V across the LDR
when the intensity is above 20 lux than the
resistor must be less than 750Ω, so it must
be decreased.
Decreased (ANS)
Example 11 2005 Question 8
(85%)
i LED converts electrical energy into light
energy.
ii Photodiode converts light energy into
electrical energy.
Example 12 2005 Question 9
Use V = iR
 5 × 10-6 × 100
= 5 × 10-4 V (ANS)
(70%)
Example 13 2005 Question 10
(70%)
Example 14 2001 Question 4
(58%)
If the resistance is infinite, then the solar cell
doesn’t supply a current.
From the graph, when the current is zero the
voltage is
0.8 V
(ANS)
Example 15 2001 Question 5
(73%)
The maximum current (when V = 0) is
20 mA (ANS)
(from the graph)
Physics Unit 3 2015
9 Photonics
Example 16 2001 Question 6
(63%)
Power = VI
At points A and D, the power supplied is zero.
At A, I = 0, and at D, V = 0.
At B, V = 0.2, and I = 18,  P = 3.6W
At C, V = 0.5, and I = 15,  P = 7.5W
 C (ANS)
Example 17 1992 Question 1
From the graph the resistance is
30 000 
= 30k
(ANS)
The biggest problem with this question was
remembering to read the vertical axis
correctly.
Example 18 1992 Question 2
When the light level is 10 lux, the resistance
of the LDR is 30 k.
This means that the resistance of the circuit is
now 300 k.
 90% of the voltage will drop across the 270
k resistor.
 the voltage at the point x will be 10% of 6V.
 0.6 V
(ANS)
Example 19 2006 Question 8
(35%)
Modulation: The varying voltage from the
microphone is converted into a varying optical
signal.
Demodulation: The varying optical signal from
the cable is converted into a varying voltage.
Example 120 2006 Question 9
(35%)
P: B, Q: A
P: An electrical signal is converted into an
optical signal.
Q: An optical signal is converted into an
electrical signal
Example 21 2004 Sample Question 7
The LED is emitting light i.e. forward biased.
Therefore the voltage across the LED is 0.6V.
Since both elements are in series, this means
that the voltage across the 200 resistor
must be 10 - 0.6 = 9.4 V.
Therefore the current in the circuit is given by
I = 9.4/200 = 0.047
47 mA
(ANS)
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Make sure that you follow the instructions and
give your answer in mA.
Example 22 2004 Sample Question 8
The light beam is not broken, so the intensity
on the LDR is 00 lux. Therefore the of the
LDR is 100  (from the graph). There totasl
resistance in this part of the circuit is 900 +
100 = 1000 Ω. The potential drop across this
part of the circuit is 10 V (it is in parallel with
the LED). Therefore the current in the circuit
is 10/1000 = 0.01 A
 10 mA
(ANS)
Make sure that you follow the instructions and
give your answer in mA.
Example 23 2004 Sample Question 9
To find the incident light intensity we need to
know the resistance of the LDR and then use
the graph to find the intensity.
With the light beam broken the voltage across
the 900 Ω resistor is 0.01 V, Therefore the
voltage across the LDR is 10 - 0.01 = 9.99 V.
Use V = iR to find the current in the circuit.
I = 0.01/900
 I = 11 A
To determine the effective resistance of the
LDR use R = 9.99/(11 x 10-6)
 R = 0.91 M.
From the graph, the light intensity falling on
the LDR is about 1.0 lux
(Note that both scales of the graph
logarithmic).
 1 lux
(ANS)
Example 24 2004 Sample Question 10
All circuit components have some effective
resistance. So there will always be some
heating with a current passing through the
component. So, no circuit component like the
LED, for example, is 100% efficient in
converting electrical energy to light energy (in
this case). This eliminates responses A and
C.
The light intensity of an LED increases as,
the current increases; this eliminates
response D.
 B (ANS)
Physics Unit 3 2015
9 Photonics
Example 25 2004 Sample Question 11
The shape must look like the original voltage
waveform, but remain positive at all times.
There is no such thing as negative optical
intensity, so even though the voltage across a
device can be positive or negative, the
energy transformation from electrical energy
to optical intensity means that the light
intensity can only be positive, or zero.
Example 26 2004 Pilot Question 7 (60%)
A current of 10mA through the LED required
1.5 V.
voltage across RD is 10 – 1.5 = 8.5V.
Applying Ohm’s Law V = iR,
 8.5 = 10  10-3 R
 RD = 850 Ω
(ANS)
Example 27 2004 Pilot Question 8 (60%)
The LED will still use 1.5V, therefore the
voltage across RD is not affected by reducing
the resistance of RD.
Therefore reducing the resistance of RD will
increase the current. Since V = iR, and since
V has remained the same, but R has
decreased, then ‘i’ needs to increase to
compensate.
This will lead to an increased current through
the LED which will increase the light output.
 INCREASE
(ANS)
Example 28 2004 Pilot Question 9 (80%)
The LED is not 100% efficient, so some heat
energy is produced.
 C (ANS)
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