A higher voltage DIY generator?

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Hi,

I am trying to create a DIY linear generator like http://www.creative-science.org.uk/gensimple1.html. However, I am wondering if this design i thought of will work or not or give more voltage or not.

Basically, instead of a solid coin-like NIB magnet like the one in website, i will be using a hollow tube shape NIB magnet instead (please refer to image http://www.filefactory.com/file/ah2f2f4/n/Scan10085_JPG ). I am wondering if this design, giving more surface area and thus more change in flux (?) will allow higher voltages to be produced. Because i need around 1.2 V to light an infrared LED for 1/30th of a second and want to minimize number of windings.

Thank you very much.

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@kbbkk,

From the article you referenced:

Quote:

The peak voltage generated by this little device is given by:

V = A x M x N

Where A is the cross sectional area of the can (0.0008 m²), M is the rate of change of magnetic field (ca. we need to use very strong magnets having a surface field of say 1 Tesla (see magnet info. below), so shaking it say 5 times a sec we get M = 5 Tesla / sec) and N the number of turns.
If we want an LED to light brightly we need to generate peak voltages of about 4V;

rearanging the formula allows us to estimate the number of turns:

N = V / (A x M) = 4 / (0.0008 x 5) = about 1000 turns - happy winding !

If I'm reading this correctly, the size of the magnet is not important, it's the cross section of the container holding it. If you want to reduce the number of windings, I suggest increasing the rpm of the magnet.

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Remembering my rusty physics, it's the strength of the magnetic field, not the shape of the magnet that affects the amount of current generated.

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A higher voltage DIY generator?

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