physics sensor coursework

At 50 lux:

VA        =        VB + p.d.

VA        =        1.64 + 3.11

VA        =        4.75 V

R1            =             (V1 R2)/ (V-V1)

R1            =             (4.75 × 1500)/ (10 – 4.75)

R1            =             1357.14 Ω

At 60 lux:

VA        =        VB + p.d.

VA        =        1.64 + 2.75

VA        =        4.39 V

R1            =             (V1 R2)/ (V-V1)

R1            =             (4.39 × 1500)/ (10 – 4.39)

R1            =             1173.80 Ω

Although the graphs indicate my prediction, of voltage output deceasing with increasing light intensity, is correct, I found many complications throughout the preliminary experiments.

     One, is the difficultly with changing the light intensity using a book or a material like that. It is impossible to accurately change the light intensity with just blocking off the light. Instead, a trial-and-error method is adopted, where we continually move the book back and forth until there is the correct light intensity.

     There is also the problem of maintaining the light intensity at the same level for a long enough time, so that the voltage output can be read off. Slight movement of the human hand can cause the book to block off more or less light to LDR. As a result, LDR resistance and thus, voltage output will be different. This leads to human error because the correct voltage output cannot be read off due to the rapid changes in the reading.

The solution to this problem would be to change the way in which the light intensity, acting upon the LDR, can be changed. The most suitable option may be to use a lamp. With a lamp, I can alter the light intensity by moving the lamp closer or further away from the LDR. In addition to this, the experiment will take place in a dark room, so there will be more control over the light intensity.

The other major problem that I found with my preliminary method is the use of a Wheatstone Bridge. I found it extremely difficult to set up as I often found that I connected the voltmeter the wrong way round, so I would get a negative output when there should be a positive output. On another occasion, I connected the circuit in such a way that the voltage output increased with increasing light intensity, which went against my prediction and research.

     This is why I have decided to use a normal potential divider circuit instead. It is much simpler to set up and it should give me similar values of resistance. When I tried setting up this circuit, I found I could do it with a lot more ease, and I need not spend as much time as I did with the Wheatstone Bridge arrangement.

     With this circuit, I may need to keep the external control variables such as humidity and temperature more constant though, in case they may lead to less viable results.

     I think I will also fit in an ammeter into the circuit, so that I can work out the resistance of the LDR more easily, using the equation, V = IR.

     My final design of circuit:

A safety issue to be considered is not to have too high a voltage setting on the power supply. If there is a high voltage of about 10V and there is a low resistance component in the circuit, then the component may heat up and there is the risk of a fire.

The range of values in which I was thinking of taking is too small. When I measured the light intensity outside the laboratory, there were light levels which reached 800 lux. This exceeds the capability of my original calibration curve by a lot. Therefore, I will take a range of eleven values between 0 lux and 1000 lux. Including 0 lux and in increasing intervals of one hundred lux, I will record the voltage output up to and including 1000 lux. For each light intensity, I will take three readings.

     This is my final method:

1. Set up new design of sensor circuit.
2. Turn off all the lights in the dark room.
3. Connect up the lamp and set the power supply to 10V.
4. Measure the light intensity beside LDR using a light meter.
5. If light intensity is above desired value, move the lamp away from the sensor; but if the light intensity is below desired value, then move the lamp towards the sensor.
6. Continue steps (ii.) and (iii.) until required value of light intensity is achieved.
7. Record the voltage output.
8. Repeat the procedure for other values of light intensity.
9. Using the equation, V = IR, work out the resistance of the LDR.
10. Plot a graph of light intensity against resistance.

Results

Set 1:

Set 2:

 

Set 3:

Set 1:

Set 2:

Set 3:

 

Averaged set:

        

Average V        =        (V1 + V2 + V3)/ 3

Average I        =        (I1 + I2 + I3)/ 3

Average R        =        Average V/ Average I

For example; at 0 lux:

Average V        =        (V1 + V2 + V3)/ 3

Average V        =        (13.18 + 13.21 +13.18)/ 3

Average V        =        13.19 V

Average I        =        (I1 + I2 + I3)/ 3

Average I        =        [(1.06 + 1.11 +1.07) ×10^-3]/ 3

Average I        =        1.08×10^-3 A

Average R        =        Average V/ Average I

Average R        =        13.19/ 1.08×10^-3

Average R        =        12 212.96 Ω

There is a degree of error in graph shown all graphs, but error displayed in graph below is due to averaging and to some extent, the resolution of measuring equipment (e.g. voltmeter reads to 2 decimal places).

Resolution of voltmeter = 0.01 V

Error in measurement if reading is constant = ±0.005 V

Max value        =        Max V/ Min I

Min value        =        Min V/ Max I

Max error min        =        (Error at lowest value/ Lowest value) ×100

Max error max        =        (Error at highest value/ Highest value) ×100

At 0 lux:

Max value        =        Max V/ Min I

Max value        =        (13.21 + 0.005)/ [(1.06 – 0.005) ×10^-3]

Max value        =        12 526.07 Ω

Min value        =        Min V/ Max I

Min value        =        (13.18 – 0.005)/ [(1.11 + 0.005) ×10^-3]

Min value        =        11 816.14 Ω

Error max        =        (313.11/ 12 526.07) ×100

Error max        =        2.50%

Error min        =        (396.82/ 11816.14) ×100

Error min        =        3.36%

At 100 lux:

Max value        =        Max V/ Min I

Max value        =        (9.82 + 0.005)/ [(7.68 – 0.005) ×10^-3]

Max value        =        1280.13 Ω

Min value        =        Min V/ Max I

Min value        =        (9.76 – 0.005)/ [(7.94 + 0.005) ×10^-3]

Min value        =        1227.82 Ω

Error max        =        (21.44/ 1280.13) ×100

Error max        =        1.67%

Error min        =        (30.87/ 1227.82) ×100

Error min        =        2.51%

At 200 lux:

Max value        =        Max V/ Min I

Max value        =        (8.02 + 0.005)/ [(12.57 – 0.005) ×10^-3]

Max value        =        638.68 Ω

Min value        =        Min V/ Max I

Min value        =        (7.97 – 0.005)/ [(12.65 + 0.005) ×10^-3]

Min value        =        629.40 Ω

Error max        =        (4.55/ 638.68) ×100

Error max        =        0.71%

Error min        =        (4.73/ 629.40) ×100

Error min        =        0.75%

At 300 lux:

Max value        =        Max V/ Min I

Max value        =        (7.26 + 0.005)/ [(13.22 – 0.005) ×10^-3]

Max value        =        549.75 Ω

Min value        =        Min V/ Max I

Min value        =        (7.16 – 0.005)/ [(13.29 + 0.005) ×10^-3]

Min value        =        538.17 Ω

Error max        =        (5.26/ 549.75) ×100

Error max        =        0.96%

Error min        =        (6.32/ 538.17) ×100

Error min        =        1.17%

At 400 lux:

Max value        =        Max V/ Min I

Max value        =        (6.75 + 0.005)/ [(16.34 – 0.005) ×10^-3]

Max value        =        413.53 Ω

Min value        =        Min V/ Max I

Min value        =        (6.71 – 0.005)/ [(16.44 + 0.005) ×10^-3]

Min value        =        407.72 Ω

Error max        =        (2.66/ 413.53) ×100

Error max        =        0.64%

Error min        =        (3.15/ 407.72) ×100

Error min        =        0.77%

At 500 lux:

Max value        =        Max V/ Min I

Max value        =        (5.99 + 0.005)/ [(17.32 – 0.005) ×10^-3]

Max value        =        346.23 Ω

Min value        =        Min V/ Max I

Min value        =        (5.92 – 0.005)/ [(17.51 + 0.005) ×10^-3]

Min value        =        337.71 Ω

Error max        =        (4.49/ 346.23) ×100

Error max        =        1.30%

Error min        =        (4.03/ 337.71) ×100

Error min        =        1.19%

At 600 lux:

Max value        =        Max V/ Min I

Max value        =        (5.28 + 0.005)/ [(18.05 – 0.005) ×10^-3]

Max value        =        292.88 Ω

Min value        =        Min V/ Max I

Min value        =        (5.23 – 0.005)/ [(18.07 + 0.005) ×10^-3]

Min value        =        289.07 Ω

Error max        =        (2.18/ 292.88) ×100

Error max        =        0.74%

Error min        =        (1.63/ 289.07) ×100

Error min        =        0.56%

At 700 lux:

Max value        =        Max V/ Min I

Max value        =        (5.02 + 0.005)/ [(18.14 – 0.005) ×10^-3]

Max value        =        277.09 Ω

Min value        =        Min V/ Max I

Min value        =        (4.83 – 0.005)/ [(18.33 + 0.005) ×10^-3]

Min value        =        263.16 Ω

Error max        =        (8.45/ 277.09) ×100

Error max        =        3.05%

Error min        =        (5.48/ 263.16) ×100

Error min        =        2.08%

At 800 lux:

Max value        =        Max V/ Min I

Max value        =        (4.35 + 0.005)/ [(18.67 – 0.005) ×10^-3]

Max value        =        233.32 Ω

Min value        =        Min V/ Max I

Min value        =        (4.14 – 0.005)/ [(18.92 + 0.005) ×10^-3]

Min value        =        218.49 Ω

Error max        =        (5.83/ 233.32) ×100

Error max        =        2.50%

Error min        =        (9/ 218.49) ×100

Error min        =        4.12%

At 900 lux:

Max value        =        Max V/ Min I

Max value        =        (4.16 + 0.005)/ [(18.73 – 0.005) ×10^-3]

Max value        =        222.43 Ω

Min value        =        Min V/ Max I

Min value        =        (3.89 – 0.005)/ [(18.78 + 0.005) ×10^-3]

Min value        =        206.81 Ω

Error max        =        (8.68/ 222.43) ×100

Error max        =        3.90%

Error min        =        (6.94/ 206.81) ×100

Error min        =        3.36%

At 1000 lux:

Max value        =        Max V/ Min I

Max value        =        (3.73 + 0.005)/ [(18.64 – 0.005) ×10^-3]

Max value        =        200.43 Ω

Min value        =        Min V/ Max I

Min value        =        (3.61 – 0.005)/ [(18.97 + 0.005) ×10^-3]

Min value        =        189.99 Ω

Error max        =        (4.89/ 200.43) ×100

Error max        =        2.44%

Error min        =        (5.55/ 189.99) ×100

Error min        =        2.92%

Analysis

Looking at the graph, we can see a very definite relationship between the light intensity and resistance of the LDR. This is that as the light intensity increases, the resistance of the LDR will decrease. However, the graph does not show linearity, which means that a change in either variable will not lead to the same proportional change in the other.

     The relationship can be explained in terms of the amount of charge carriers in the LDR. The charge carriers in my particular circuit are electrons. Generally, the more delocalised electrons there are in a material, the lower the resistivity. This is because the delocalised electrons will move and respond to a potential difference by carrying charge. Thus, the more delocalised electrons there are, the more charge that can be carried, and the lower the resistance.

     An LDR is made out of a semi conducting material, which means that it has electrical properties intermediate to that of conductors and insulators. However, the resistivity of semi conductors decreases with increasing temperature. This is as the heat energy allows more electrons to break free from their atoms, and hence join the conduction band.

     The same can be applied to the light intensity. The higher the light intensity, the more light photons hit the LDR. These light photons carry packets of energy, which are transferred to the atoms in the LDR, and the electrons can therefore use this energy to become delocalised. With more delocalised electrons, the LDR has more charge carriers, and thus, there is a lower resistance.

     In addition to this, a positive hole is created when an electron jumps into the conduction band. This can also move and carry charge, as the atoms with the hole steal other atom’s electrons, in an attempt to gain a full valence band.

     Although the atoms also vibrate more with the energy, which can impede the flow of electrons, the effect of having additional delocalised electrons is more than enough to compensate for this.

     This change is not directly proportional though. This is because as more electrons break free, the less localised electrons are left in the LDR. Therefore, a further increase in energy will cause fewer electrons to break free, and thus a smaller decrease in resistance. The effect is seen by the reduction in rate of change, as the light intensity rises.

     Therefore, it is possible to assume the resistance will increase again after a certain level. This is because when the maximum possible amount of electrons has broken free, the amount of charge carriers will not change any more. However, the atomic vibrations can still increase though, which can increase the resistance in the LDR.

     The results correspond with my original prediction and preliminary research, of resistance decreasing with increasing light intensity.

        

In the averaged graph shown above, there is only one anomaly, which is the value at 200 lux. This can be caused by a variety of factors. These may involve a change in the external environment, such as the humidity in the surroundings or the temperature.

     It is most likely that the effect is caused by an alteration in the temperature. As explained above, temperature can have an effect on the number of delocalised electrons in the conductance band of LDR. Therefore, a change in temperature may lead to there being extra electrons.

     Looking at how my averaged reading is less than predicted, it is likely that on the day I did the experiment and got this reading, the temperature was higher than usual.

To analyse how well the sensor performed in our circuit though, there are several criteria to consider. These include the resolution, sensitivity, response time, unsystematic fluctuations, systematic error and linearity.

     The resolution is the smallest detectable detail in a sensor. When performing the experiment, the LDR proved to have a very high resolution. Even a slight change in light intensity could lead to a change in the voltage output. For example, if we were to leave the lamp in a particular position, the voltage output may still change slightly every now and then. This is probably caused by the dust in the atmosphere, which may block off some of the light photons.

     However, another possibility is the introduction of unsystematic fluctuations, which is noise. Unsystematic fluctuations can be found in all sensors and measuring systems. They are usually where the values of voltage or current changes randomly and unexpectedly. There is a bit of noise in our circuit, as the readings never stay exactly the same for long.

     The sensitivity of a sensor is the ratio of change of output to input. As the graph does not show linearity, the sensitivity of the sensor is different at various light intensities. The change in sensitivity is very rapid though. Before 100 lux, we experience a large change in resistance even with a small change in light intensity. After 300 lux though, any additional change in light intensity causes an extremely small change in the resistance. This can be measured as the gradient of the graph. For example, the sensitivity of the sensor at 40 lux is approximately 200 Ω lux-1; but at 600 lux, the sensitivity is less than 0.5 Ω lux-1.

     The response time is the time it takes for the resistance of the LDR to change when there is a change in light intensity. The LDR’s resistance is relatively good, but sometimes it can take over 1s before the LDR reacts to the change. However, this is not important in our particular circuit, so the sensor remains fairly well-suited to our circuit.

     Something that I noticed was that there were voltages of over 13 V, which was physically impossible when I only set the power supply to 10 V. Therefore, I tried finding the true EMF of the power supply by connecting a voltmeter directly across the power supply. The voltmeter gave a reading of 13.96 V. The extra 3.96 V would be systematic error, as the error would be in all readings derived from this power supply.

E        =        8.15 V

I        =        0.08 A

R        =        100 Ω

E        =        IR + Ir

r        =        (E/I) – R

r        =        (8.15/0.08) – 100

r        =        1.9 Ω

Internal resistance in the power supply may also cause systematic error. An experiment to discover the value of the internal resistance was done. This gave us a reading of less than 2 Ω, so it is negligible. It would already be unnecessary because the voltage output reading would already have experienced the drop in voltage due to internal resistance. Therefore, internal resistance does not need to be taken in account.

     The final aspect to consider is the linearity of the circuit. This may produce some problems when producing a calibration graph, because a simple equation cannot be derived.

Deriving an equation

We can convert the graph from a curve into a straight line by using a logarithmic scale instead of a linear scale:

(3, 2.3) (0, 4.7)                m = -0.8, y-intercept = 4.7

R        =        50118.72336 × L-0.8

(2.3, 3) (4.7, 0)        m = -1.25, y-intercept = 5.87

L        =        749894.2093 × R-1.25

To use this equation, you must know the voltage output and the current in the circuit. These values allow the resistance of the LDR to be calculated. Then, substitute the resistance into equation and work out for light intensity. I tested the calibration equation outside the laboratory. Locations that I did this at include a corridor and beside a window.

     In the corridor, V = 9.84 V, I = 7.56 × 10^-3 A, L = 96 lux.

Using my formula:

R        =        V/ I

R        =        9.84/ (7.56 × 10^-3)

R        =        1301.59 Ω

L        =        749894.2093 × R-1.25

L        =        749894.2093 × 1301.59-1.25

L        =        95.92 lux

The true light intensity and my calculated value are extremely similar, with a difference of only 0.08 lux.

     Beside a window, V = 4.92, I = 18.57 × 10^-3 A, L = 723 lux.

Using my formula:

R        =        V/ I

R        =        4.92/ (18.57 × 10^-3)

R        =        264.94 Ω

L        =        749894.2093 × R-1.25

L        =        749894.2093 × 264.94-1.25

L        =        701.55 lux

The difference between the true light intensity and that which was calculated was a bit larger than the previous test. Still, my equation is still relatively accurate, with an error of less than 25 lux in both cases above.

Evaluation

        

The procedure in which I have designed and carried out throughout this investigation was fairly reliable, as it provided me with results which matched my prediction and preliminary research.

     The accuracy in which the investigation was carried out is relatively high, as I used measuring equipment which could measure the voltage output and current in circuit to an accuracy of up to 2 decimal places. If you were to look at my averaged graph, there are error bars. The error bars show that there is greatest error when the light intensity was extremely low. As the light intensity increased, the error that existed became so small that it was negligible. The calculations of maximum percentage error also support this; the highest value of maximum percentage error was 4.12% at 800 lux.

Still, there was one anomaly. The cause of this anomaly is believed to be derived from a change in an external variable, which I think is the surrounding temperature. The possible effect in which this could have on the voltage output and current in circuit is described and explained in the analysis.

     The procedure was very suitable for the investigation, considering the limitations of performing it within a school. Although there is an anomalous result, it will not hamper the reliability of the investigation greatly because it was not far off the expected result. This means that the conclusion would still be valid.

     The reliability of my two equations is not extremely high even though its accuracy is more than adequate for the two locations I have tested it at. This is because I have only tested it at two locations. To improve its reliability, I should test it at several more locations, preferably with different values of light intensity.

The reliability of the results could be further improved if the external factors such as temperature could be more strongly controlled. This can be achieved if there was some sort of thermostat in the dark room. The thermostat could be set to a particular level, and be maintained at that level until the experiment was complete.

     In addition to this, there would be some error because of the internal heating of the LDR. This would have a greater effect on the results rather than the external change in temperature. To solve this, I could leave the LDR to cool down for a minute in between each reading.

     Another change to the procedure which I would propose would be to link the circuit to a computer, so a computer could record the results. The reason for this is that the readings on the voltmeter and ammeter change constantly due to noise, so it is impossible to interpret an accurate reading just by inspection. In addition to this, the light intensity can sometimes be very difficult to change if we were to do it by hand, moving the lamp closer or further away from the LDR. If this were to be done by a computer, the computer could increase the power supplied to the lamp so that its brightness increases or decreases.

     Both of these improvements would enable us to gather more accurate results. The first proposal of maintaining the temperature at constant level might be possible within a school. However, the second alteration would be impractical because it would be very difficult for the school to get a computer or suitable software which can perform the operations mentioned.

Safety precautions that should be taken into account include wearing rubber gloves. This is in case there is some damage to the insulating layer of the wires, which can lead to an electric shock.

     Also, the temperature of the LDR and the resistor should be constantly monitored. Current passing through a component with any resistance will result in heat being produced. Therefore, if the voltage setting on the power supply was too high, the resistors may overheat, which can lead to a fire.

My proposed new method:

1. Set up sensor circuit.
2. Turn off all the lights in the dark room.
3. Set 23˚C on the thermostat, and leave the room at that temperature for 5 minutes.
4. Connect up the lamp and set the power supply to 10V.
5. Measure the light intensity beside LDR using a light meter.
6. If light intensity is above desired value, move the lamp away from the sensor; but if the light intensity is below desired value, then move the lamp towards the sensor.
7. Continue steps (ii.) and (iii.) until required value of light intensity is achieved.
8. Record the voltage output.
9. Turn power supply off for a minute, so the circuit can cool down.
10. Repeat the procedure for other values of light intensity.
11. Using the equation, V = IR, work out the resistance of the LDR.
12. Plot a graph of light intensity against resistance.
13. Plot a graph of log light intensity against log resistance.
14. Derive a formula linking the two variables using y = kxn.
15. Test the formula in a minimum of six different locations.

Bibliography