Investigating the effect of intensity on the power of solar cells.

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Investigating the effect of intensity on the power of solar cells

This experiment involves changing the intensity of light falling on different cells and measuring their power outputs. Higher intensity of light means that there are more photons hitting the surface of the cell per unit area per second. The more hit the cell, the more rapidly the electrons move across the p-n junction, so the larger the emf produced. If the rate of movement of electrons is inhibited, then the greater the rate of supply of photons (intensity), the more will not successfully excite an electron, so the lower the efficiency of the cell. I therefore predict that the higher the intensity, the greater the emf across the cell, the greater the power output of the cell, and the lower the efficiency of the cell.

Method

The intensity of the light falling on the cell will vary with the separation of the light source and the cell according to the inverse square law:

intensity  1/d2 so I can vary the intensity by changing the distance between the light source and the cell. I have calculated the amount of light hitting the cell by the ratio of area of the cell to the ratio of area over which light is spread:

intensity = power from bulb x surface area of cell

                                                            4 π r2

The power from the bulb was worked out from the current through the bulb and voltage put across it by the power pack. An estimated efficiency of 2% was used.

Outline: I varied the distance of the light source from the cell, then took readings from a voltmeter in parallel, and an ammeter in series, with the solar cell.

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 Rough trials showed the following:

        Using the thin layer polycrystalline cell, I found that pressing contacts onto the surface of the cell scratched it, so I taped wire pickups onto the top and bottom with transparent tape, as shown on the pictures. I noticed that the readings obtained for the current produced by the cell were inconsistent between ammeters (see Experiment Five). I therefore changed my method for the amorphous and monocrystalline cells and used the voltage readings and the resistance of the voltmeter to work out the power output of the cell by P = V2/R.

        Having ...

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