physics sensor coursework

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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. If the potential at terminal A is below that of terminal B, the voltage output will be negative. If the potential at terminal A is greater than that of terminal B, caused by a decrease in light intensity, the voltage output will be positive. As a result, my prediction is that as the light intensity increases, the voltage output will decrease.

Methodology

The investigation will be carried out in the laboratory. This is so that external factors – except for the independent variable – remain relatively constant. Such factors that need to be controlled include the temperature of components and humidity of external environment, although the Wheatstone Bridge arrangement already overcomes these.

     I will take a range of seven values between 0 lux and 60 lux. Including 0 lux and in increasing intervals of ten lux, I will record the voltage output up to and including 60 lux. For each light intensity, I will take two readings.

The equipment involved in this investigation includes;

  • Power supply – To supply power to the lamp and to the sensor circuit.
  • LDR – This is the sensor in which I will be calibrating.
  • Fixed resistors ×3 – These are so the potential is divided across components of different resistance.
  • Voltmeter – To measure the voltage across the resistor.
  • Opaque material (e.g. a book) – To block off some light, allowing alteration of light intensity.
  • Light meter – To measure the light levels that falls upon the LDR.
  • Amplifier – To increase the sensitivity of circuit.

The circuit shall be set up exactly as above. Using some sort of opaque material, I shall be varying the light intensity acting on LDR by blocking off some of the light. The reading from the voltmeter can then be recorded.

     This is my preliminary method:

  1. Set up sensor circuit.
  2. Keep light intensity within laboratory at a constant level (≈100 lux).
  3. Measure the light intensity beside LDR using a light meter.
  4. If light intensity is above desired value, move the opaque object towards the sensor, in order to block off more light; but if the light intensity is below desired value, move the opaque object away from the sensor, so more light falls on the LDR.
  5. Continue steps (ii.) and (iii.) until required value of light intensity is achieved.
  6. Record the voltage output.
  7. Repeat the procedure for other values of light intensity.

 

Preliminary results

I worked out the resistance by first finding the constant potential across terminal B, using the potential divider equation:

V1        =        (V R1)/ (R1 + R2)

V1        =        (10 × 1000)/ (1000 + 5100)

V1        =        1.64 V

Then I worked out the voltage across the LDR at terminal A, by adding on the voltage difference to the potential across terminal B:

p.d.        =        VA - VB

VA        =        VB + p.d.

VA        =        1.64 + p.d.

Finally I can work out the resistance by rearranging the potential divider equation:

V1        =        (V R1)/ (R1 + R2)

V1 (R1 + R2)        =        V R1 

V1 R1+ V1R2          =             V R1

V1 R2         =        V R1 - V1 R1

V1 R2         =         R1 (V - V1)

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

At 0 lux:

VA        =        VB + p.d.

VA        =        1.64 + 7.23

VA        =        8.87 V

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

R1            =             (8.87 × 1500)/ (10 – 8.87)

R1            =             11 774.34 Ω

At 10 lux:

VA        =        VB + p.d.

VA        =        1.64 + 6.31

VA        =        7.95 V

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

R1            =             (7.95 × 1500)/ (10 – 7.95)

R1            =             5817.07 Ω

At 20 lux:

VA        =        VB + p.d.

VA        =        1.64 + 4.75

VA        =        6.39 V

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

R1            =             (6.39 × 1500)/ (10 – 6.39)

R1            =             2655.12 Ω

At 30 lux:

VA        =        VB + p.d.

VA        =        1.64 + 3.87

VA        =        5.51 V

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

R1            =             (5.51 × 1500)/ (10 – 5.51)

R1            =             1840.76 Ω

At 40 lux:

VA        =        VB + p.d.

VA        =        1.64 + 3.47

VA        =        5.11 V

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

R1            =             (5.11 × 1500)/ (10 – 5.11)

R1            =             1567.48 Ω

Join now!

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 ...

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