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# physics sensor coursework

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Introduction

Physics coursework – sensor circuits

In this investigation, I will be constructing a circuit which involves a sensor. I will calibrate this circuit and then test it in an environment outside the laboratory.

The sensor I have chosen to use is a light dependant resistor, LDR. From my preliminary research, I found that this sensor works by changing its resistance as light levels are modified. This change is not proportional though, but a curve as shown:  The resistance changes because the LDR is made up of a semi conducting material. When more photons hit the LDR, more electrons break free to act as charge carriers, thus reducing resistance in material.

Circuit

I shall incorporate the LDR into a potential divider circuit, in an arrangement called a “Wheatstone Bridge”. An advantage of the Wheatstone Bridge, which is described in the “AS Advancing Physics” textbook, is that it only detects changes in the ratio of the two pairs of resistors. This means that an amplifier would only increase the changes in voltage output, rather than the whole thing. Also, the circuit would be unaffected by any outside changes – such as temperature – as the variable would cause the same proportional effect to each resistor. This means the ratio of each pair of resistors remains constant, so same output voltage.

Above is the preliminary design of my circuit. Terminal A is connected to the positive; whilst terminal B is connected to the negative. It works because an increase in light intensity causes the LDR’s resistance to drop. This drop in resistance means there is less potential at terminal A in the circuit. Potential at terminal B stays the same as the resistors are fixed. The potential difference across the two terminals is measured.

Middle

267.36

800

4.35

18.67

232.99

900

3.89

18.73

207.69

1000

3.73

18.85

197.88

Set 2:

 Light intensity/ lux Potential difference/ V Current/ mA Resistance/ ohms 0 13.21 1.11 11 900.90 100 9.76 7.69 1269.18 200 7.97 12.65 630.04 300 7.26 13.27 547.10 400 6.73 16.34 411.87 500 5.99 17.51 342.09 600 5.28 18.05 292.52 700 4.83 18.25 264.66 800 4.32 18.72 230.77 900 3.98 18.77 212.04 1000 3.61 18.97 190.30

Set 3:

 Light intensity/ lux Potential difference/ V Current/ mA Resistance/ ohms 0 13.18 1.07 12 317.76 100 9.76 7.94 1229.22 200 8.02 12.58 637.52 300 7.16 13.22 541.60 400 6.75 16.44 410.58 500 5.92 17.32 341.80 600 5.23 18.06 289.59 700 5.02 18.33 273.87 800 4.14 18.92 218.82 900 4.16 18.78 221.51 1000 3.70 18.64 198.50

Set 1:  Set 2:  Set 3:  Averaged set:

 Light intensity/ lux Potential difference/ V Current/ mA Resistance/ ohms 0 13.19 1.08 12 212.96 100 9.78 7.77 1258.69 200 7.99 12.60 634.13 300 7.22 13.26 544.49 400 6.73 16.38 410.87 500 5.96 17.44 341.74 600 5.25 18.06 290.70 700 4.90 18.24 268.64 800 4.27 18.77 227.49 900 4.01 18.76 213.75 1000 3.68 18.82 195.54

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]

Conclusion

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

 Source Information http://www.tiscali.co.uk/reference/ encyclopaedia/hutchinson/m0030304.html LDR is a semi conductor, made out of a material such as cadmium sulphide. http://en.wikipedia.org/wiki/Photocell LDR conducts electricity via positive holes and negative electrons. http://www.electroflash.org.nz/schoolh/symbolsall.htm The symbol for an LDR. http://www.technologystudent.com/elec1/ldr1.htm Resistance falls rapidly from 0 lux to 100 lux. http://www.gcsescience.com/pe27.htm Graph of light intensity against resistance for an LDR. AS Advancing Physics – Edited by Jon Ogborn and Mary Whitehouse The advantages of using a Wheatstone bridge over a normal circuit.

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