Chapter 5 Introducing Photonics

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Heinemann Physics 12(3e)Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk(a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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Worked solutionsChapter 5 Introducing photonics

5.1 Photonics in telecommunications

1 B. Using light to manipulate information and energy. Photonics is not just about

telecommunications.

2 B. The optical spectrum includes those wavelengths that are either visible or near visible so as

to be usable in manipulating information, e.g. infrared remote controls.

3 A. Optical fibres can handle many thousands times more information than copper.

4 D. Depth of penetration decreases as the frequency of the signal increases for copper. High-

frequency signals therefore propagate inefficiently, low frequencies efficiently. In optical fibres

the loss does not depend on frequency.

5 D. From answer to Question 4, high-frequency signals propagate inefficiently in copper wire.

6 A. The photophone was invented by Alexander Graham Bell in 1880, 4 years after he invented

the telephone.

7 D. 95.5% over 1 km is close to the theoretical limit of transmission through glass fibre.

8 C. The Internet was developed by the US military as a distributed system to counter a potential

attack on centralised communication systems.

9 A. Bell lacked the glass fibres and coherent light sources that modern optical systems rely on.



Worked solutions Chapter 5 Introducing photonics


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5.2 Optical transducers

1 A. The intensity of the light emitted by an LED is directly proportional to the forward-biased

current flowing through the p-n junction.

2 C. An LD is a laser diode. It operates on the same principle as an LED but over a much

narrower wavelength due to a special current-confining geometry.

3 Advantage: higher optical gain; disadvantage: generally slower response time

4 This is a voltage divider circuit and is solved accordingly.

a In series Rtotal = RLDR + Rout = 2 M + 6.8 k = 2 M

b V out = V t ! Rout/[ RLDR + Rout])

=

3

3 6

9 6.8 10

6.8 10 2 10

! !

! + !

= 0.03 V

c If V out = 6 V then V LDR = 3 VThe ratio of the resistances will be the same as the voltages.

RLDR = V LDR /V out ! Rout =3

3

6 6.8 10! !

= 3.4 ! 10 3 or 3.4 k

5 This is a voltage divider circuit and is solved accordingly. Refer also to Figure 5.13, page 156,

in the text. a If V out = 4 V then V R = 12 – 4 = 8 V and RLED = V out/V R ! RR = 4/8 ! 20 = 10 k

Reading incident light intensity from Figure 5.13 for this value gives an intensity of

500–550 mW m –2

, which is above the minimum acceptable level.

.

b From part a: 500–550 mW m –2

6 a V R = 50 ! 106 ! 50 ! 10

–9 = 2.5 V

Hence V R d = 3.5 V

b Photoconduction mode (since the diode is reverse biased)

7 a I = V / R = 10/500 ! 103 = 2 ! 10

–5 = 20 A since I must be the same for all loads in a

series circuit.

b Max. intensity corresponding to 20 ! 10 –6

A from Figure 5.16, p. 159, is 4 W m –2

8 At minimum operating level V L = V c – V LED = 9 – 2 = 7 V

RL = V R / I = 3

9 2

20 10!

!

" = 350

9 a Circuit i: In parallel, required current = 3 ! 20 mA = 60 mA

V L = V C – V LED = 9 – 2 = 7 V

RL = V R / I = 7/60 ! 10 –3

= 117

Circuit ii: In series, required current = 20 mA, V L = V C – 3 ! V LED = 9 – 6 = 3 V

RL = V R / I = = 3/20 ! 10 –3

= 150

b Both circuits will emit the same light—all are operating at same V and I . c Circuit ii, requires only 20 mA, so will run longer.

10 a From the graph, RT = 500 at 20°C





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b V T = 10 – 2.5 = 7.5 V, I T = 7.5/500 = 0.015 A

since I LED = 0.011, I R = 0.004 A and RR = 2.5/0.004 = 625

c RLED = 2.0/0.0048 = 416.7

Hence I R = 416.7/625 ! 0.0048 = 0.0032 A and I T = 0.0032 + 0.0048 = 0.008 A

V T = 10 – 2.0 = 8.0 V

Hence RT = 8.0/0.008 = 1000

Using this value on the graph supplied, T = 10 °C





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5.3 Audio transmission via a light beam

1 A. Intensity modulation

2 B. Dispersion of an LD is less hence better suited to long distances

3 D. Options A, B and C all apply to the choice of a phototransistor

4 B. Attachment of the device limits speakers response to high frequencies

5 C. Outer barrel of the casing is the positive terminal.

6 D. Output varies between 2.9 and 4.5 V.

7 A. 12 V exceeds operating voltage.

8 A. Outer case is positive terminal so must be insulated.

9 B. Any excess supply can damage the laser diode.

10 C. Capacitor is acting as a filter.





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Chapter review

1 B. From theory, photodiodes should be reverse biased.

2 D. From theory, photodiodes have a faster response than phototransistors which are faster thanLDRs.

3 B. From theory, a reverse-biased laser diode would be the most suitable.

4 a P = VI = 2 ! 100 mA = 200 mW = 0.2 W

b Efficiency = output/input ! 100 = 0.002/0.2 ! 100 = 1%

5 Active area = "r 2 = "(2 !10

–3)

2 = 1.26 !10

–5 m

2

#= P / A =4

5

1 10

1.26 10

!

!

"

"

= 4 W m –2

6 V R 1 = 10 – 3 = 7 V

Hence circuit I = 7/50 ! 103

= 0.14 mA

Hence RLED = 3/1.4 ! 10 –4

= 21.4 k

Reading from Figure 5.13, page 156, intensity = 130 mW m –2

7 I L = 6/300 ! 103 = 0.02 mA or 20 ! 10

-6A

Reading from Figure 5.16, page 159, intensity = 4 W m –2

8 a If intensity increases, photocurrent increases (see Figure 5.16, page 159), R1 decreases

and hence V 1 decreases. V 2 must therefore increase for V T = 9 V.

b From Figure 5.16, page 159, at intensity = 3 W m –2

, I = 15 ! 10

–6A.

Light intensity per unit area = 30 mW m –2. From Figure 5.13, RLDR = approx. 43 kHence V LDR = 15 ! 10

-6! 43 ! 10

3= approx. 0.64 V

c R2 = R1 for V 2 = V 1 so R2 = 43 k

d No. Any other light intensity will increase/decrease V LDR and hence decrease/increase V 2.

Only at one particular value will V 1 = V 2

9 a RLDR = 86 k

b V out will increase to the maximum V for the circuit = 10 V

c V out will decrease to approx. 0 V.





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Exam-style questions (Electronics andphotonics)

1 a 2.0 A

b The reading on the voltmeter is equal to the terminal voltage of the battery, i.e. 216 V.

c 216 V

d 440 W

e 432 W

2 220 V

3 a 98 $

b 1.0 A

c i 80 V

ii 10 Viii 8.0 V

d 98 V

e 100 W

4 a 60 $

b 2.0 A

c I 1 = 1.20 A, I 2 = 0.60 A, I 3 = 0.20 A

d 240 W

e 240 W

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R 1 ( ) R 2 ( ) V out (V)1000 1000 10

3000 1000 5.0

400 100 4.0

900 100 2.0

2.0 3.0 12.0

400 100 4.0

6 a 0.5

b 2

c 4.0 ! 10 –18

Jd 8.0 ! 10

–18J

e 100 V





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7

8 a A semiconductor is a material which has a greater resistance than metals, but less than

insulators. Silicon and germanium are examples. It conducts as a result of electrons

gaining sufficient energy to break out of the crystal lattice. This may result from an

energy input from light or heat.

b n-type semiconductors are doped with an element such as phosphorus which has 5

electrons in its outer shell (silicon has 4) and thus the extra one is mobile and becomes a

charge carrier. p-type semiconductors are doped with an element such as aluminiumwhich has 3 electrons in its outer shell (silicon has 4) and thus the missing electron acts as

a ‘hole’ into which mobile electrons can jump, thus resulting in the hole moving and

becoming a charge carrier.

c A pn junction is the interface between a p-type and n-type semiconductor where the extra

electrons and holes will tend to recombine, forming a ‘depletion layer’ which does not

conduct unless forward biased.

9 If the p-type material in a pn junction is made positive relative to the n-type material, holes will

flow towards the junction from the p-type and electrons from the n-type, thus they recombine

and a current flows. Thus the pn junction is said to be forward biased. If the p-type is negative

and n-type positive, that is the junction is reverse biased, the holes and electrons withdraw

further from the junction, thus making it effectively an insulator.

10 a 20 mA

b 20 mW

c 200 "

d 0.10 W

e 20 mA

11 a 1.5 V

b 0.150 W

c 0

d 3.5 V

e 0 W

f 35 "

g 0.35 W

h 0.50 W

12 a 1.0 V

b 20 mW

c 0

13 a 22 mA

b 11 mA

c 11 mA





8

14 a 309.1"

b 0.15 W

c 220 mW

15 150 "

16 A diode will only conduct electricity when it is connected in forward bias. When a diode is

connected in reverse bias, its resistance is so great as to produce only a negligible current.

17 The voltage across the terminals of a diode beyond which the diode conducts strongly.

For silicon V s – 0.7 V; for germanium V s – 0.3 V.

18 1.0 ! 10 –3

C

19 When the capacitor is fully charged, no more current flows in the circuit so no voltage drop

across the resistor and all of the voltage across the battery appears across the capacitor.

V c = Q/C =6

9

50 10

100 10

!

!

"

"

= 500 V

20 It is an inverting amplifier because the amplifier output voltage has the opposite sign to the

input voltage (over the linear amplification range). For example, V in = +0.5 V, V out = –10 V and

for V in = –0.50 V, V out = +10 V.

21 a 0.50 V

b –20 V

22

Peak input voltage Output voltage range100 mV –2 to +2 V

0.25 V –5 to +5 V

1.0 V –20 to +20 V

20 mV –400 to +400 mV

0.80 V –16 to +16 V

23 a This is a non-inverting amplifier. This is because a positive input voltage becomes a

positive output voltage, and a negative input voltage becomes a negative output voltage.

b 53

c 5.3 V

24 a A thermistor is a temperature-sensitive resistor whose resistance decreases as its

temperature increases. It is usually constructed from Ge, Si or a mixture of various

oxides. The resistance of a typical thermistor may range from 10 k " at 0ºC to 100 "

at 100ºC.

b





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25 The coupling capacitor isolates (or blocks) any DC component of the collector voltage of the

first stage from the input to the second stage (but passes the AC signal voltage). This means that

the DC biasing conditions of the second stage are not disturbed by the DC voltage of the first

stage collector.

26 a 4 V

b Above; V out rises.

c V out approaches zero.

27 a 200 "

b 87 "

c 79 mA

d 2.6 V

28 a A photodiode is like a normal diode but the pn junction is exposed to light which can

release electron-hole pairs when the diode is reverse biased, thus allowing a small reverse

current which is proportional to the amount of light to flow.b The dark current is the current that flows when the reverse biased while no light falls on

the diode. It is very small, well under a microamp.

c The current through a photodiode is only zero when there is a small positive voltage

across it to oppose the photocurrent.

29 a 500 k "

b 32 W

c 150 W

d 50 W

e 200 W

30 A phototransistor is a transistor that has a photosensitive collector–base pn junction. The current produced by the photoelectric effect due to incident light is the base current of this device. The

result of this is that an increase in the incident light intensity will produce an increase in

collector current.

31 a 9.0 W

b 3.71 ! 10 –19

J

c 2.42 ! 1013

d 3.88 A

e 51.5

32 a An LED is a diode that will emit visible light when it is energised. When the diode is

forward biased the recombination of holes and electrons takes place. During this

recombination, some of the excess energy possessed by the unbound free electrons in the

pn junction is converted into light energy.

b From the table, the threshold voltage is around 1.6 V.

c 1.6 k "

d 22 mW

e 100 mW

33 E

Chapter 5 Introducing Photonics

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