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Physics Unit 3 2015 9 Photonics 1 of 25 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. Physics Unit 3 2015 9 Photonics 2 of 25 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. Physics Unit 3 2015 9 Photonics 3 of 25 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 Physics Unit 3 2015 9 Photonics 4 of 25 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. Physics Unit 3 2015 9 Photonics 5 of 25 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. Physics Unit 3 2015 9 Photonics 6 of 25 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 10A. 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. Physics Unit 3 2015 9 Photonics 7 of 25 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. Physics Unit 3 2015 9 Photonics 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. 8 of 25 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? Physics Unit 3 2015 9 Photonics 9 of 25 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. Physics Unit 3 2015 9 Photonics 10 of 25 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? Physics Unit 3 2015 9 Photonics 11 of 25 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. Physics Unit 3 2015 9 Photonics 12 of 25 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? Physics Unit 3 2015 9 Photonics 13 of 25 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? Physics Unit 3 2015 9 Photonics 14 of 25 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. Physics Unit 3 2015 9 Photonics 15 of 25 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. Physics Unit 3 2015 9 Photonics 16 of 25 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. Physics Unit 3 2015 9 Photonics 17 of 25 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) Physics Unit 3 2015 9 Photonics 18 of 25 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. Physics Unit 3 2015 9 Photonics 19 of 25 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. Physics Unit 3 2015 9 Photonics 20 of 25 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. Physics Unit 3 2015 9 Photonics 21 of 25 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. Physics Unit 3 2015 9 Photonics 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) 22 of 25 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) Physics Unit 3 2015 9 Photonics 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 23 of 25 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) 24 of 25 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) 25 of 25