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Most bicycle hub dynamos (generators) output about 6V. Are there 12V (or 13.5V) versions? Has anyone worked out a converter for using the 6V output of the standard dynamos to charge a 12V lead acid battery?

(based on a weak memory from late '60's) The light set on my bike had a three position switch. On battery (2 Ds), Off, and On generator. Rewiring so light would be on from generator while moving, and stay on from batteries at stops without switching, ate batteries and gave dim light while moving.<p>Working backwards from that questionable info:
The dynamo was probably 3VAC RMS.<p>If they are AC, a transformer can be used to step-up the voltage; and provides isolation from bike frame. (with isolation, a fullwave bridge rectifier can be used.)<p>I didn't find a standard transformer. Maybe a 115V to 24V backwards, remove turns from 115V side till you get about 16V out after the rectifier.

the output should be speed related so just ride faster!

If it is just a dynamo (And not also a built in rectifier) and its output is AC then Dale's suggestion should work but you will also half the current therby doubling the charge time. But since its frequency is variable with velocity, the transformer may not work efficiently at some speeds.<p>If you take joesmiths observation that the output V is proportional to the speed of the input shaft then you can put a smaller roller on the generator which will make it spin faster for the same bike velosity. there will be a limit because too small a roller will slip on the wheel. Perhaps you can rig a gearing system to increase input revs. <p>Finally you can modify the generator by adding more turns in it's windings or increase the strength of it's permenant magnet. both easier said than done.

them dynomos are ac output...and they are sealed,, no way to open them up without destroying them,,,
my suggestion is to find you a way to use a dc motor in someway and let the wheel drive it..
besure to use a regulator..as hi speed = hi volts
you do not want to fry anything when going down hill..lol
good luck

Another way to do this is a rube goldburg. Use the 6 volts to power a 6v motor. This drives a 12 volt motor which outputs 12 volts(actually slightly less due to loss). Speeds still need to be matched but its now an electic transmission so to speak. Just step the speed up a bit with gears or choose the rated speeds for volts.<p>OR buy 2- 6V motors and drive the 2nd twice as fast. This might ruin it though.

Use the older type stepper motor out of the full height 5 1/4 inch floppys. <p>They crank out more than what you need, and that means more current.

It sounds like the basic generator is an alternator with a permanent magnet field. I suspect the the magnet rotates and that the windings are stationary. This avoids the use of slip rings, which would be more expensive to build and less reliable.<p>The open circuit voltage will be a function of speed, and will probably vary over a wide range. The frequency would also vary. However, the integral of voltage over a cycle will be constant. The decrease in time compensates for the increase in voltage.<p>The flux change in a transformer also depends on the integral of voltage over a cycle. Thus a transformer which works at one combination of voltage and speed should work at all combinations of voltage and speed.<p>The battery charge voltage will fall within a narrow range around 14 volts. If the generator output was rectified and applied directly to the battery, there would be no charging until the battery voltage was exceeded. Above this point, the current flow would increase rapidly, possibly to a level that would damage the generator.<p>Some type of control circuit between the generator and battery would be helpful. A switching circuit is probably best, for reasons of efficiency. At low speeds, it should draw maximum output from the grnerator. The input impedance of the regulator should match the internal impedance of the generator. The generator output voltage would be loaded down to one-half of its open circuit voltage.<p>At some point, the current will reach the maximum current rating of the generator. This current should be maintained as the generator voltage increases further. This current should be reduced if necessary to control the battery charging. rate.<p>A flyback converter is probably appropriate for this application, since it tends to deliver constant energy to the load, and allows the load to assume its own voltage level, especially if current mode control is used.<p>Stepping the voltage up before rectifying it would probably help the efficiency. A 6 volt generator output could loose about 10% of its output in the rectifier. Schottky barrier diodes would help a little. Doubling the voltage to 12 volts with a transformer would reduce the rectifier loss by half for either rectifier type.<p>[ May 04, 2004: Message edited by: stephen ]</p>

Thank for the replies!

To keep a battery charged, does the charger have to exceed the battery voltage by a certain margin?

Go to your local autozone store and buy an alternator for a 79 chevy for $19<p> I don't know what I'm talking about <p>

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by keymaker:
Go to your local autozone store and buy an alternator for a 79 chevy for $19<p> I don't know what I'm talking about <p>
<hr></blockquote><p>
Yeah maybe at 1000 rpms. LOL

Don't you know that all you have to do is use a very small pulley and ride down hill?<p>

If you have a 6 volt AC output, could you use a voltage doubler to achieve 12 volts and still have sufficient current to charge your battery? Seems like it might be worth a try.