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Voltage produced by the bicycle dynamo varies depending on what speed your traveling at. I have two 12 volt bulbs mounted on the bike at the front and at the rear. I want to power these bulbs using a 12 volt battery until the dynamo produces a voltage greater than 12 volts then I want to switch to the dynamo to power the lights.

How would I do this?

Update

For those interested this is the circuit I built. http://dcaclab.com/en/experiments/7662-battery-and-dynamo-bicycle-circuit/

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    \$\begingroup\$ Are you wanting to charge the battery with the dynamo? \$\endgroup\$
    – markrages
    May 2 '12 at 19:36
  • \$\begingroup\$ Are you talking pedal bike or motor bike? \$\endgroup\$
    – jippie
    May 2 '12 at 19:43
  • \$\begingroup\$ @jippie it is a pedal bike. \$\endgroup\$
    – Neil
    May 2 '12 at 19:53
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    \$\begingroup\$ @markrages Perhaps in the future I'll charge the battery with the dynamo. But for now I'm just using a battery that can't be recharged. \$\endgroup\$
    – Neil
    May 2 '12 at 19:53
  • \$\begingroup\$ @Neil - Do you have a datasheet (or at least a manufacturer and model) for this dynamo? Typical bike dynamos put out 6V RMS of Alternating Current and will never reach 12V. \$\endgroup\$ May 2 '12 at 21:10
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All you need is a single power diode if it is a rechargable battery, two diodes if the battery can't be recharged.

In the first case, the battery is always connected to the light. If the dynamo voltage is high enough, then it will dump current onto the battery-light combination. This current will charge the battery minus whatever is used to run the light. This setup is probably only good for a lead-acid battery, which are pretty forgiving in how they are charged. I'm also making the assumption that the dynamo can't put out enough power to overcharge the battery, which is probably true for a normal bicycle dynamo and even a small lead-acid battery.

If the battery can't be recharged, then you have to make sure the dynamo can't force reverse current thru the battery. In that case you need two diodes, one from the battery to the light and the other from the dynamo to the light. In this setup, whichever of the dynamo or the battery is at higher voltage will power the light.

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  • \$\begingroup\$ Thanks I might try the rechargeable circuit in the future. For now I'm using a battery that can't be recharged. Would it be possible to use a LED as the diode? That way I could visible see which power source is being used to power the lights just by looking at the circuit. \$\endgroup\$
    – Neil
    May 2 '12 at 19:48
  • \$\begingroup\$ no, a LED won't work because the current you run through it would be much higher than 20mA, won't it? \$\endgroup\$ May 2 '12 at 20:10
  • \$\begingroup\$ An LED won't work both because your 12V lamp will likely require too much current as Noah1989 points out (unless the lamp is an LED array itself) and because few LEDs can withstand a reverse voltage of more than about 5V, see the answers to this question for more information and check the datasheet of your LED. \$\endgroup\$ May 2 '12 at 20:25
  • \$\begingroup\$ In case the voltage drop from the diodes is too much for you, you might want to look for ideal diode circuits which basically use a MOSFET to emulate a diode. \$\endgroup\$
    – ndim
    Jan 17 '13 at 21:56
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Two diodes will do what you want. Diode from battery to the lights, diode from dynamo to the lights. The highest voltage will supply the current.

That said, powering light bulbs with higher than rated voltage is bad for them. You might consider some switching regulator to convert the higher voltage to 12V.

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  • \$\begingroup\$ A switching regulator isn't necessary. The bulb always draws the same current at 12V, and reducing the current drawn from the high-voltage dynamo output won't decrease the pedal resistance meaningfully. A simple linear regulator should be fine. \$\endgroup\$ May 2 '12 at 20:09
  • \$\begingroup\$ A watt's a watt. A casual cyclist is putting out a couple hundred watts total. And those dynamos aren't terribly efficient, so burning a watt of output wastes several watts on the input side. \$\endgroup\$
    – markrages
    May 2 '12 at 20:42
  • \$\begingroup\$ The dynamos aren't terribly efficient, no, but they aren't linear either. According to the graphs at this page, the output power of all tested dynamos actually increases when you drop the output current from 250 mA to a lower value. \$\endgroup\$ May 2 '12 at 21:06

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